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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? If Thomas were a doctor, then it is overcast. If it snows, then Thomas is a doctor. If it is overcast, then William is an electrician. If William were an electrician, then it is overcast. | If Thomas were a doctor, then William is an electrician. If it snows, then it is overcast. It is overcast if and only if William is an electrician. William is an electrician if and only if it is overcast. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? If Thomas were a doctor, then it is overcast. If it snows, then Thomas is a doctor. If it is overcast, then William is an electrician. If William were an electrician, then it is overcast.
### Response:
If Thomas were a doctor, then William is an electrician. If it snows, then it is overcast. It is overcast if and only if William is an electrician. William is an electrician if and only if it is overcast.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: There is at least one z for which if z were an office clerk, then z is a sailor. For all z, if z were a sailor, then z is an office clerk. For all z, z will not go running. For all z, z will go running if and only if z is a sailor. | There is at least one z for which z is an office clerk if and only if z is a sailor can be inferred via the existential biconditional introduction rule. There is at least one z for which z is a sailor if and only if z is an office clerk can be inferred via the existential biconditional introduction rule. For all z, if z goes running, then z is a sailor can be inferred via the universal biconditional elimination rule. For all z, if z were a sailor, then z is running can be inferred via the universal biconditional elimination rule. For all z, z is not a sailor can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: There is at least one z for which if z were an office clerk, then z is a sailor. For all z, if z were a sailor, then z is an office clerk. For all z, z will not go running. For all z, z will go running if and only if z is a sailor.
### Response:
There is at least one z for which z is an office clerk if and only if z is a sailor can be inferred via the existential biconditional introduction rule. There is at least one z for which z is a sailor if and only if z is an office clerk can be inferred via the existential biconditional introduction rule. For all z, if z goes running, then z is a sailor can be inferred via the universal biconditional elimination rule. For all z, if z were a sailor, then z is running can be inferred via the universal biconditional elimination rule. For all z, z is not a sailor can be inferred via the universal biconditional elimination rule.
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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x: P(x). | Nothing can be inferred from these premises. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x: P(x).
### Response:
Nothing can be inferred from these premises.
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Consider the following premises. p_10 <-> p_22. ~q. s <-> q. p_22 -> ~q_22. q_22. Can we infer s from them? If possible, name the inference rules being used at each step. | From p_22 -> ~q_22, q_22 we can infer ~p_22 via modus tollens. Finally, from s <-> q, ~q we can infer ~s via biconditional elimination, which contradicts s. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. p_10 <-> p_22. ~q. s <-> q. p_22 -> ~q_22. q_22. Can we infer s from them? If possible, name the inference rules being used at each step.
### Response:
From p_22 -> ~q_22, q_22 we can infer ~p_22 via modus tollens. Finally, from s <-> q, ~q we can infer ~s via biconditional elimination, which contradicts s. Therefore, the answer is no.
| null | null |
Consider the following premises. exists x_9: S_2(x_9) <-> S(x_9). forall x_9: ~P_2(x_9). forall x_9: P_2(x_9) or S(x_9). Can we infer forall x_9: ~S_2(x_9) from them? If possible, name the inference rules being used at each step. | From exists x_9: S_2(x_9) <-> S(x_9) we can infer exists x_9: S(x_9) -> S_2(x_9) via existential biconditional elimination. From forall x_9: P_2(x_9) or S(x_9), forall x_9: ~P_2(x_9) we can infer forall x_9: S(x_9) via universal disjunctive syllogism. Finally, from exists x_9: S(x_9) -> S_2(x_9), forall x_9: S(x_9) we can infer exists x_9: S_2(x_9) via existential modus ponens, which contradicts forall x_9: ~S_2(x_9). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. exists x_9: S_2(x_9) <-> S(x_9). forall x_9: ~P_2(x_9). forall x_9: P_2(x_9) or S(x_9). Can we infer forall x_9: ~S_2(x_9) from them? If possible, name the inference rules being used at each step.
### Response:
From exists x_9: S_2(x_9) <-> S(x_9) we can infer exists x_9: S(x_9) -> S_2(x_9) via existential biconditional elimination. From forall x_9: P_2(x_9) or S(x_9), forall x_9: ~P_2(x_9) we can infer forall x_9: S(x_9) via universal disjunctive syllogism. Finally, from exists x_9: S(x_9) -> S_2(x_9), forall x_9: S(x_9) we can infer exists x_9: S_2(x_9) via existential modus ponens, which contradicts forall x_9: ~S_2(x_9). Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~q_16. p_16 or q_16. w <-> p. w -> ~r_16. ~p_16 or r_16. | q_16 or r_16 can be inferred via the resolution rule. w -> p can be inferred via the biconditional elimination rule. p -> w can be inferred via the biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~q_16. p_16 or q_16. w <-> p. w -> ~r_16. ~p_16 or r_16.
### Response:
q_16 or r_16 can be inferred via the resolution rule. w -> p can be inferred via the biconditional elimination rule. p -> w can be inferred via the biconditional elimination rule.
| null | null |
Consider the following premises. forall x_1: Q(x_1) -> T_2(x_1). forall x_1: Q(x_1) <-> T_2(x_1). exists x_1: T_2(x_1) -> Q(x_1). Can we infer exists x_1: Q(x_1) -> T_2(x_1) from them? | From forall x_1: Q(x_1) <-> T_2(x_1) we can infer forall x_1: T_2(x_1) -> Q(x_1). From exists x_1: T_2(x_1) -> Q(x_1), forall x_1: Q(x_1) -> T_2(x_1) we can infer exists x_1: T_2(x_1) <-> Q(x_1). Finally, from exists x_1: T_2(x_1) <-> Q(x_1) we can infer exists x_1: Q(x_1) -> T_2(x_1). Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_1: Q(x_1) -> T_2(x_1). forall x_1: Q(x_1) <-> T_2(x_1). exists x_1: T_2(x_1) -> Q(x_1). Can we infer exists x_1: Q(x_1) -> T_2(x_1) from them?
### Response:
From forall x_1: Q(x_1) <-> T_2(x_1) we can infer forall x_1: T_2(x_1) -> Q(x_1). From exists x_1: T_2(x_1) -> Q(x_1), forall x_1: Q(x_1) -> T_2(x_1) we can infer exists x_1: T_2(x_1) <-> Q(x_1). Finally, from exists x_1: T_2(x_1) <-> Q(x_1) we can infer exists x_1: Q(x_1) -> T_2(x_1). Therefore, the answer is yes.
| null | null |
Consider the following premises. forall x_11: P_11(x_11) or Q_11(x_11). forall x_11: ~P_11(x_11) or R_11(x_11). forall x_11: W(x_11). exists x_11: Q_11(x_11) -> T_4(x_11). forall x_11: R_11(x_11) -> T_4(x_11). Can we infer exists x_11: T_4(x_11) and W(x_11) from them? If possible, name the inference rules being used at each step. | From forall x_11: P_11(x_11) or Q_11(x_11), forall x_11: ~P_11(x_11) or R_11(x_11) we can infer forall x_11: Q_11(x_11) or R_11(x_11) via universal resolution. From exists x_11: Q_11(x_11) -> T_4(x_11), forall x_11: R_11(x_11) -> T_4(x_11), forall x_11: Q_11(x_11) or R_11(x_11) we can infer exists x_11: T_4(x_11) via existential disjunction elimination. Finally, from exists x_11: T_4(x_11), forall x_11: W(x_11) we can infer exists x_11: T_4(x_11) and W(x_11) via existential conjunction. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_11: P_11(x_11) or Q_11(x_11). forall x_11: ~P_11(x_11) or R_11(x_11). forall x_11: W(x_11). exists x_11: Q_11(x_11) -> T_4(x_11). forall x_11: R_11(x_11) -> T_4(x_11). Can we infer exists x_11: T_4(x_11) and W(x_11) from them? If possible, name the inference rules being used at each step.
### Response:
From forall x_11: P_11(x_11) or Q_11(x_11), forall x_11: ~P_11(x_11) or R_11(x_11) we can infer forall x_11: Q_11(x_11) or R_11(x_11) via universal resolution. From exists x_11: Q_11(x_11) -> T_4(x_11), forall x_11: R_11(x_11) -> T_4(x_11), forall x_11: Q_11(x_11) or R_11(x_11) we can infer exists x_11: T_4(x_11) via existential disjunction elimination. Finally, from exists x_11: T_4(x_11), forall x_11: W(x_11) we can infer exists x_11: T_4(x_11) and W(x_11) via existential conjunction. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: There is at least one x_7 for which if x_7 reads a book, then x_7 is climbing a mountain. For all x_7, x_7 will not read a book. For all x_7, if x_7 climbs a mountain, then x_7 is a bartender. For all x_7, x_7 will read a book or x_7 is not a bartender. | There is at least one x_7 for which if x_7 reads a book, then x_7 is a bartender can be inferred via the existential transitivity rule. For all x_7, if x_7 were a bartender, then x_7 is reading a book can be inferred via the universal rewriting rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: There is at least one x_7 for which if x_7 reads a book, then x_7 is climbing a mountain. For all x_7, x_7 will not read a book. For all x_7, if x_7 climbs a mountain, then x_7 is a bartender. For all x_7, x_7 will read a book or x_7 is not a bartender.
### Response:
There is at least one x_7 for which if x_7 reads a book, then x_7 is a bartender can be inferred via the existential transitivity rule. For all x_7, if x_7 were a bartender, then x_7 is reading a book can be inferred via the universal rewriting rule.
| null | null |
Consider the following premises. forall x_8: S_2(x_8) <-> R(x_8). forall x_8: T(x_8) -> S_2(x_8). forall x_8: T(x_8) or R(x_8). exists x_8: U(x_8). Can we infer forall x_8: ~(S_2(x_8) and U(x_8)) from them? | From forall x_8: S_2(x_8) <-> R(x_8) we can infer forall x_8: R(x_8) -> S_2(x_8). From forall x_8: T(x_8) -> S_2(x_8), forall x_8: R(x_8) -> S_2(x_8), forall x_8: T(x_8) or R(x_8) we can infer forall x_8: S_2(x_8). Finally, from forall x_8: S_2(x_8), exists x_8: U(x_8) we can infer exists x_8: S_2(x_8) and U(x_8), which contradicts forall x_8: ~(S_2(x_8) and U(x_8)). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_8: S_2(x_8) <-> R(x_8). forall x_8: T(x_8) -> S_2(x_8). forall x_8: T(x_8) or R(x_8). exists x_8: U(x_8). Can we infer forall x_8: ~(S_2(x_8) and U(x_8)) from them?
### Response:
From forall x_8: S_2(x_8) <-> R(x_8) we can infer forall x_8: R(x_8) -> S_2(x_8). From forall x_8: T(x_8) -> S_2(x_8), forall x_8: R(x_8) -> S_2(x_8), forall x_8: T(x_8) or R(x_8) we can infer forall x_8: S_2(x_8). Finally, from forall x_8: S_2(x_8), exists x_8: U(x_8) we can infer exists x_8: S_2(x_8) and U(x_8), which contradicts forall x_8: ~(S_2(x_8) and U(x_8)). Therefore, the answer is no.
| null | null |
Consider the following premises. forall x_13: S(x_13) <-> Q_13(x_13). exists x_13: P_7(x_13) -> S(x_13). forall x_13: P_7(x_13) or Q_13(x_13). forall x_13: S(x_13) -> U_1(x_13). Can we infer forall x_13: ~U_1(x_13) from them? If possible, name the inference rules being used at each step. | From forall x_13: S(x_13) <-> Q_13(x_13) we can infer forall x_13: Q_13(x_13) -> S(x_13) via universal biconditional elimination. From exists x_13: P_7(x_13) -> S(x_13), forall x_13: Q_13(x_13) -> S(x_13), forall x_13: P_7(x_13) or Q_13(x_13) we can infer exists x_13: S(x_13) via existential disjunction elimination. Finally, from forall x_13: S(x_13) -> U_1(x_13), exists x_13: S(x_13) we can infer exists x_13: U_1(x_13) via existential modus ponens, which contradicts forall x_13: ~U_1(x_13). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_13: S(x_13) <-> Q_13(x_13). exists x_13: P_7(x_13) -> S(x_13). forall x_13: P_7(x_13) or Q_13(x_13). forall x_13: S(x_13) -> U_1(x_13). Can we infer forall x_13: ~U_1(x_13) from them? If possible, name the inference rules being used at each step.
### Response:
From forall x_13: S(x_13) <-> Q_13(x_13) we can infer forall x_13: Q_13(x_13) -> S(x_13) via universal biconditional elimination. From exists x_13: P_7(x_13) -> S(x_13), forall x_13: Q_13(x_13) -> S(x_13), forall x_13: P_7(x_13) or Q_13(x_13) we can infer exists x_13: S(x_13) via existential disjunction elimination. Finally, from forall x_13: S(x_13) -> U_1(x_13), exists x_13: S(x_13) we can infer exists x_13: U_1(x_13) via existential modus ponens, which contradicts forall x_13: ~U_1(x_13). Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_9: ~S(x_9). exists x_9: S_1(x_9) -> Q_9(x_9). forall x_9: S(x_9) <-> Q_9(x_9). | forall x_9: S(x_9) -> Q_9(x_9). forall x_9: Q_9(x_9) -> S(x_9). forall x_9: ~Q_9(x_9). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_9: ~S(x_9). exists x_9: S_1(x_9) -> Q_9(x_9). forall x_9: S(x_9) <-> Q_9(x_9).
### Response:
forall x_9: S(x_9) -> Q_9(x_9). forall x_9: Q_9(x_9) -> S(x_9). forall x_9: ~Q_9(x_9).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? p_1 -> R_7(a_7). R_7(a_7) -> p_1. forall x_7: ~R_7(x_7). | Nothing can be inferred from these premises. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? p_1 -> R_7(a_7). R_7(a_7) -> p_1. forall x_7: ~R_7(x_7).
### Response:
Nothing can be inferred from these premises.
| null | null |
Consider the following premises. forall x_7: R_3(x_7) <-> Q_7(x_7). forall x_7: W(x_7) or R_3(x_7). forall x_7: Q_1(x_7) -> Q_7(x_7). forall x_7: W(x_7) -> Q_1(x_7). Can we infer forall x_7: Q_7(x_7) from them? If possible, name the inference rules being used at each step. | From forall x_7: R_3(x_7) <-> Q_7(x_7) we can infer forall x_7: R_3(x_7) -> Q_7(x_7) via universal biconditional elimination. From forall x_7: W(x_7) -> Q_1(x_7), forall x_7: Q_1(x_7) -> Q_7(x_7) we can infer forall x_7: W(x_7) -> Q_7(x_7) via universal transitivity. Finally, from forall x_7: W(x_7) -> Q_7(x_7), forall x_7: R_3(x_7) -> Q_7(x_7), forall x_7: W(x_7) or R_3(x_7) we can infer forall x_7: Q_7(x_7) via universal disjunction elimination. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_7: R_3(x_7) <-> Q_7(x_7). forall x_7: W(x_7) or R_3(x_7). forall x_7: Q_1(x_7) -> Q_7(x_7). forall x_7: W(x_7) -> Q_1(x_7). Can we infer forall x_7: Q_7(x_7) from them? If possible, name the inference rules being used at each step.
### Response:
From forall x_7: R_3(x_7) <-> Q_7(x_7) we can infer forall x_7: R_3(x_7) -> Q_7(x_7) via universal biconditional elimination. From forall x_7: W(x_7) -> Q_1(x_7), forall x_7: Q_1(x_7) -> Q_7(x_7) we can infer forall x_7: W(x_7) -> Q_7(x_7) via universal transitivity. Finally, from forall x_7: W(x_7) -> Q_7(x_7), forall x_7: R_3(x_7) -> Q_7(x_7), forall x_7: W(x_7) or R_3(x_7) we can infer forall x_7: Q_7(x_7) via universal disjunction elimination. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: s_4 -> r_1. p_8 <-> s_4. w_4 -> p_8. | p_8 -> s_4 can be inferred via the biconditional elimination rule. s_4 -> p_8 can be inferred via the biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: s_4 -> r_1. p_8 <-> s_4. w_4 -> p_8.
### Response:
p_8 -> s_4 can be inferred via the biconditional elimination rule. s_4 -> p_8 can be inferred via the biconditional elimination rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: exists z: S(z) <-> Q(z). forall z: ~S(z). forall z: Q(z) <-> S(z). | forall z: Q(z) -> S(z) can be inferred via the universal biconditional elimination rule. exists z: S(z) -> Q(z) can be inferred via the existential biconditional elimination rule. forall z: S(z) -> Q(z) can be inferred via the universal biconditional elimination rule. exists z: Q(z) -> S(z) can be inferred via the existential biconditional elimination rule. exists z: ~Q(z) can be inferred via the existential biconditional elimination rule. forall z: ~Q(z) can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: exists z: S(z) <-> Q(z). forall z: ~S(z). forall z: Q(z) <-> S(z).
### Response:
forall z: Q(z) -> S(z) can be inferred via the universal biconditional elimination rule. exists z: S(z) -> Q(z) can be inferred via the existential biconditional elimination rule. forall z: S(z) -> Q(z) can be inferred via the universal biconditional elimination rule. exists z: Q(z) -> S(z) can be inferred via the existential biconditional elimination rule. exists z: ~Q(z) can be inferred via the existential biconditional elimination rule. forall z: ~Q(z) can be inferred via the universal biconditional elimination rule.
| null | null |
Translate the following inference to logic notation: For all x, if x_9 reads a book, then x_9 is fast. There is at least one x for which x_9 will not read a book. For all x, if x_9 were fast, then x_9 is reading a book. For all x, x_9 is excited if and only if x_9 is fast. Therefore there is at least one x for which x_9 is not excited. | forall x: P(x) -> Q(x). exists x: ~P(x). forall x: Q(x) -> P(x). forall x: R(x) <-> Q(x). Therefore exists x: ~R(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, if x_9 reads a book, then x_9 is fast. There is at least one x for which x_9 will not read a book. For all x, if x_9 were fast, then x_9 is reading a book. For all x, x_9 is excited if and only if x_9 is fast. Therefore there is at least one x for which x_9 is not excited.
### Response:
forall x: P(x) -> Q(x). exists x: ~P(x). forall x: Q(x) -> P(x). forall x: R(x) <-> Q(x). Therefore exists x: ~R(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? ~u_3. ~q_3 -> w. s <-> s_2. s <-> u_3. q_3 -> s_2. | s -> s_2. s -> u_3. s_2 -> s. u_3 -> s. ~s. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? ~u_3. ~q_3 -> w. s <-> s_2. s <-> u_3. q_3 -> s_2.
### Response:
s -> s_2. s -> u_3. s_2 -> s. u_3 -> s. ~s.
| null | null |
Translate the following inference to logic notation: There is at least one x for which x_8 is not an artist or x_8 will take a plane. For all x, x_8 is an electrician or x_8 is a musician. For all x, if x_8 were a musician, then x_8 is an artist. For all x, x_8 is an artist or x_8 is not an electrician. Therefore there is at least one x for which x_8 will make tea or x_8 will take a plane. | exists x: ~P(x) or Q(x). forall x: R(x) or S(x). forall x: S(x) -> P(x). forall x: P(x) or ~R(x). Therefore exists x: T(x) or Q(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: There is at least one x for which x_8 is not an artist or x_8 will take a plane. For all x, x_8 is an electrician or x_8 is a musician. For all x, if x_8 were a musician, then x_8 is an artist. For all x, x_8 is an artist or x_8 is not an electrician. Therefore there is at least one x for which x_8 will make tea or x_8 will take a plane.
### Response:
exists x: ~P(x) or Q(x). forall x: R(x) or S(x). forall x: S(x) -> P(x). forall x: P(x) or ~R(x). Therefore exists x: T(x) or Q(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_3, x_3 will make tea and x_3 is a politician. | For all x_3, x_3 will make tea. For all x_3, x_3 is a politician. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_3, x_3 will make tea and x_3 is a politician.
### Response:
For all x_3, x_3 will make tea. For all x_3, x_3 is a politician.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Elisabeth is not a scientist. For all x_14, x_14 is a scientist and x_14 will play a game. | Since the premises are contradictory, we can infer anything from them. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Elisabeth is not a scientist. For all x_14, x_14 is a scientist and x_14 will play a game.
### Response:
Since the premises are contradictory, we can infer anything from them.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? If Joseph were a scientist, then it is snowing. It is not overcast. If it snows, then Joseph is a scientist. Joseph is a scientist if and only if it is overcast. | Joseph is a scientist if and only if it is snowing. It is snowing if and only if Joseph is a scientist. If Joseph were a scientist, then it is overcast. If it is overcast, then Joseph is a scientist. Joseph is not a scientist. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? If Joseph were a scientist, then it is snowing. It is not overcast. If it snows, then Joseph is a scientist. Joseph is a scientist if and only if it is overcast.
### Response:
Joseph is a scientist if and only if it is snowing. It is snowing if and only if Joseph is a scientist. If Joseph were a scientist, then it is overcast. If it is overcast, then Joseph is a scientist. Joseph is not a scientist.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_9, if x_9 writes a letter, then x_9 is making tea. For all x_9, x_9 will make tea if and only if x_9 will play tennis. For all x_9, if x_9 makes tea, then x_9 is writing a letter. | For all x_9, x_9 will write a letter if and only if x_9 will make tea. For all x_9, x_9 will make tea if and only if x_9 will write a letter. For all x_9, if x_9 makes tea, then x_9 is playing tennis. For all x_9, if x_9 plays tennis, then x_9 is making tea. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_9, if x_9 writes a letter, then x_9 is making tea. For all x_9, x_9 will make tea if and only if x_9 will play tennis. For all x_9, if x_9 makes tea, then x_9 is writing a letter.
### Response:
For all x_9, x_9 will write a letter if and only if x_9 will make tea. For all x_9, x_9 will make tea if and only if x_9 will write a letter. For all x_9, if x_9 makes tea, then x_9 is playing tennis. For all x_9, if x_9 plays tennis, then x_9 is making tea.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_17: P(x_17) <-> Q_17(x_17). forall x_17: P(x_17) or R_2(x_17). exists x_17: ~Q_17(x_17). | forall x_17: P(x_17) -> Q_17(x_17) can be inferred via the universal biconditional elimination rule. forall x_17: Q_17(x_17) -> P(x_17) can be inferred via the universal biconditional elimination rule. exists x_17: ~P(x_17) can be inferred via the existential biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_17: P(x_17) <-> Q_17(x_17). forall x_17: P(x_17) or R_2(x_17). exists x_17: ~Q_17(x_17).
### Response:
forall x_17: P(x_17) -> Q_17(x_17) can be inferred via the universal biconditional elimination rule. forall x_17: Q_17(x_17) -> P(x_17) can be inferred via the universal biconditional elimination rule. exists x_17: ~P(x_17) can be inferred via the existential biconditional elimination rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall z_1: ~S(z_1) or U_1(z_1). forall z_1: S(z_1) or P(z_1). forall z_1: T(z_1). forall z_1: ~P(z_1). | forall z_1: P(z_1) or U_1(z_1). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall z_1: ~S(z_1) or U_1(z_1). forall z_1: S(z_1) or P(z_1). forall z_1: T(z_1). forall z_1: ~P(z_1).
### Response:
forall z_1: P(z_1) or U_1(z_1).
| null | null |
Consider the following premises. forall x_6: ~P_6(x_6) or T(x_6). forall x_6: P_3(x_6) <-> W(x_6). forall x_6: P_6(x_6) or W(x_6). forall x_6: T(x_6) -> P_3(x_6). Can we infer exists x_6: ~P_3(x_6) from them? If possible, name the inference rules being used at each step. | From forall x_6: P_6(x_6) or W(x_6), forall x_6: ~P_6(x_6) or T(x_6) we can infer forall x_6: W(x_6) or T(x_6) via universal resolution. From forall x_6: P_3(x_6) <-> W(x_6) we can infer forall x_6: W(x_6) -> P_3(x_6) via universal biconditional elimination. Finally, from forall x_6: W(x_6) -> P_3(x_6), forall x_6: T(x_6) -> P_3(x_6), forall x_6: W(x_6) or T(x_6) we can infer forall x_6: P_3(x_6) via universal disjunction elimination, which contradicts exists x_6: ~P_3(x_6). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_6: ~P_6(x_6) or T(x_6). forall x_6: P_3(x_6) <-> W(x_6). forall x_6: P_6(x_6) or W(x_6). forall x_6: T(x_6) -> P_3(x_6). Can we infer exists x_6: ~P_3(x_6) from them? If possible, name the inference rules being used at each step.
### Response:
From forall x_6: P_6(x_6) or W(x_6), forall x_6: ~P_6(x_6) or T(x_6) we can infer forall x_6: W(x_6) or T(x_6) via universal resolution. From forall x_6: P_3(x_6) <-> W(x_6) we can infer forall x_6: W(x_6) -> P_3(x_6) via universal biconditional elimination. Finally, from forall x_6: W(x_6) -> P_3(x_6), forall x_6: T(x_6) -> P_3(x_6), forall x_6: W(x_6) or T(x_6) we can infer forall x_6: P_3(x_6) via universal disjunction elimination, which contradicts exists x_6: ~P_3(x_6). Therefore, the answer is no.
| null | null |
Consider the following premises. For all x_2, if x_2 were a poet, then x_2 is poor. There is at least one x_2 for which x_2 is not poor. For all x_2, if x_2 were poor, then x_2 is rich. For all x_2, if x_2 were rich, then x_2 is a poet. Can we infer the following from them? There is at least one x_2 for which x_2 is not a poet. | From the fact that for all x_2, if x_2 were poor, then x_2 is rich, and that for all x_2, if x_2 were rich, then x_2 is a poet we can infer that for all x_2, if x_2 were poor, then x_2 is a poet. From the fact that for all x_2, if x_2 were a poet, then x_2 is poor, and that for all x_2, if x_2 were poor, then x_2 is a poet we can infer that for all x_2, x_2 is a poet if and only if x_2 is poor. Finally, from the fact that for all x_2, x_2 is a poet if and only if x_2 is poor, and that there is at least one x_2 for which x_2 is not poor we can infer that there is at least one x_2 for which x_2 is not a poet. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_2, if x_2 were a poet, then x_2 is poor. There is at least one x_2 for which x_2 is not poor. For all x_2, if x_2 were poor, then x_2 is rich. For all x_2, if x_2 were rich, then x_2 is a poet. Can we infer the following from them? There is at least one x_2 for which x_2 is not a poet.
### Response:
From the fact that for all x_2, if x_2 were poor, then x_2 is rich, and that for all x_2, if x_2 were rich, then x_2 is a poet we can infer that for all x_2, if x_2 were poor, then x_2 is a poet. From the fact that for all x_2, if x_2 were a poet, then x_2 is poor, and that for all x_2, if x_2 were poor, then x_2 is a poet we can infer that for all x_2, x_2 is a poet if and only if x_2 is poor. Finally, from the fact that for all x_2, x_2 is a poet if and only if x_2 is poor, and that there is at least one x_2 for which x_2 is not poor we can infer that there is at least one x_2 for which x_2 is not a poet. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_6, if x_6 were a nurse, then x_6 is not an astronaut. For all x_6, x_6 will work if and only if x_6 is a nurse. For all x_6, x_6 is an astronaut. There is at least one x_6 for which x_6 will go running if and only if x_6 will work. | For all x_6, x_6 is not a nurse. For all x_6, if x_6 works, then x_6 is a nurse. There is at least one x_6 for which if x_6 goes running, then x_6 is working. For all x_6, if x_6 were a nurse, then x_6 is working. There is at least one x_6 for which if x_6 works, then x_6 is running. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_6, if x_6 were a nurse, then x_6 is not an astronaut. For all x_6, x_6 will work if and only if x_6 is a nurse. For all x_6, x_6 is an astronaut. There is at least one x_6 for which x_6 will go running if and only if x_6 will work.
### Response:
For all x_6, x_6 is not a nurse. For all x_6, if x_6 works, then x_6 is a nurse. There is at least one x_6 for which if x_6 goes running, then x_6 is working. For all x_6, if x_6 were a nurse, then x_6 is working. There is at least one x_6 for which if x_6 works, then x_6 is running.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Thomas is not playing a game. It is cloudy or Patricia is not making tea. If Thomas plays a game, then it is cloudy. If it is cloudy, then Thomas is playing a game. | It is not cloudy. If Patricia makes tea, then it is cloudy. Thomas is playing a game if and only if it is cloudy. It is cloudy if and only if Thomas is playing a game. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Thomas is not playing a game. It is cloudy or Patricia is not making tea. If Thomas plays a game, then it is cloudy. If it is cloudy, then Thomas is playing a game.
### Response:
It is not cloudy. If Patricia makes tea, then it is cloudy. Thomas is playing a game if and only if it is cloudy. It is cloudy if and only if Thomas is playing a game.
| null | null |
Consider the following premises. For all x_13, if x_13 were a police officer, then x_13 is making tea. For all x_13, if x_13 plays tennis, then x_13 is making tea. For all x_13, x_13 will read a book and x_13 will listen to a song. There is at least one x_13 for which x_13 will play tennis or x_13 is a police officer. Can we infer the following from them? For all x_13, either x_13 will read a book or x_13 will make tea are not true. | That contradicts the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_13, if x_13 were a police officer, then x_13 is making tea. For all x_13, if x_13 plays tennis, then x_13 is making tea. For all x_13, x_13 will read a book and x_13 will listen to a song. There is at least one x_13 for which x_13 will play tennis or x_13 is a police officer. Can we infer the following from them? For all x_13, either x_13 will read a book or x_13 will make tea are not true.
### Response:
That contradicts the premises. Therefore the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? p_8 <-> q_8. q -> p_8. p -> q. | p -> p_8. p_8 -> q_8. q_8 -> p_8. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? p_8 <-> q_8. q -> p_8. p -> q.
### Response:
p -> p_8. p_8 -> q_8. q_8 -> p_8.
| null | null |
Consider the following premises. ~t. p and r. t <-> s. Can we infer s from them? If possible, name the inference rules being used at each step. | That contradicts the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. ~t. p and r. t <-> s. Can we infer s from them? If possible, name the inference rules being used at each step.
### Response:
That contradicts the premises. Therefore the answer is no.
| null | null |
Translate the following inference to logic notation: For all x, if y drives a car, then y is listening to a song. For all x, if y reads a book, then y is listening to a song. For all x, if y were a politician, then y is not listening to a song. There is at least one x for which y will read a book. Therefore there is at least one x for which y is not a politician. | forall x: P(x) -> Q(x). forall x: R(x) -> Q(x). forall x: S(x) -> ~Q(x). exists x: R(x). Therefore exists x: ~S(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, if y drives a car, then y is listening to a song. For all x, if y reads a book, then y is listening to a song. For all x, if y were a politician, then y is not listening to a song. There is at least one x for which y will read a book. Therefore there is at least one x for which y is not a politician.
### Response:
forall x: P(x) -> Q(x). forall x: R(x) -> Q(x). forall x: S(x) -> ~Q(x). exists x: R(x). Therefore exists x: ~S(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_8, x_8 will go running. For all x_8, x_8 will cook if and only if x_8 will go running. | For all x_8, if x_8 cooks, then x_8 is running. For all x_8, if x_8 goes running, then x_8 is cooking. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_8, x_8 will go running. For all x_8, x_8 will cook if and only if x_8 will go running.
### Response:
For all x_8, if x_8 cooks, then x_8 is running. For all x_8, if x_8 goes running, then x_8 is cooking.
| null | null |
Consider the following premises. forall x_10: R_10(x_10) -> S_10(x_10). r_1 or ~p_4. r_1 -> R_10(a_10). Can we infer S_10(a_10) from them? If possible, name the inference rules being used at each step. | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_10: R_10(x_10) -> S_10(x_10). r_1 or ~p_4. r_1 -> R_10(a_10). Can we infer S_10(a_10) from them? If possible, name the inference rules being used at each step.
### Response:
We cannot infer that from the premises. Therefore the answer is no.
| null | null |
Translate the following inference to logic notation: If Robert were a police officer, then Mary is listening to a song. For all x, if x_10 listens to a song, then x_10 is an artist. If Mary were an artist, then Susan is making tea. Therefore if Robert were a police officer, then Susan is making tea. | p -> P(a). forall x: P(x) -> Q(x). Q(a) -> q. Therefore p -> q. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: If Robert were a police officer, then Mary is listening to a song. For all x, if x_10 listens to a song, then x_10 is an artist. If Mary were an artist, then Susan is making tea. Therefore if Robert were a police officer, then Susan is making tea.
### Response:
p -> P(a). forall x: P(x) -> Q(x). Q(a) -> q. Therefore p -> q.
| null | null |
Consider the following premises. It is late. Michael is a police officer. Can we infer the following from them? If we can, name the inference rule being used: It is late. | That is one of the premises. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. It is late. Michael is a police officer. Can we infer the following from them? If we can, name the inference rule being used: It is late.
### Response:
That is one of the premises. Therefore, the answer is yes.
| null | null |
Translate the following inference to logic notation: There is at least one x for which x_15 is a carpenter if and only if x_15 is an artist. For all x, if x_15 were a carpenter, then x_15 is a doctor. For all x, x_15 is an artist if and only if x_15 is a doctor. Therefore there is at least one x for which if x_15 were a carpenter, then x_15 is an artist. | exists x: P(x) <-> Q(x). forall x: P(x) -> R(x). forall x: Q(x) <-> R(x). Therefore exists x: P(x) -> Q(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: There is at least one x for which x_15 is a carpenter if and only if x_15 is an artist. For all x, if x_15 were a carpenter, then x_15 is a doctor. For all x, x_15 is an artist if and only if x_15 is a doctor. Therefore there is at least one x for which if x_15 were a carpenter, then x_15 is an artist.
### Response:
exists x: P(x) <-> Q(x). forall x: P(x) -> R(x). forall x: Q(x) <-> R(x). Therefore exists x: P(x) -> Q(x).
| null | null |
Translate the following inference to logic notation: For all x, if x_6 goes running, then x_6 is a soldier. For all x, x_6 is a soldier or x_6 is not an office clerk. For all x, x_6 will go running or x_6 is an office clerk. For all x, x_6 is a police officer. Therefore for all x, x_6 is a soldier and x_6 is a police officer. | forall x: P(x) -> Q(x). forall x: Q(x) or ~R(x). forall x: P(x) or R(x). forall x: S(x). Therefore forall x: Q(x) and S(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, if x_6 goes running, then x_6 is a soldier. For all x, x_6 is a soldier or x_6 is not an office clerk. For all x, x_6 will go running or x_6 is an office clerk. For all x, x_6 is a police officer. Therefore for all x, x_6 is a soldier and x_6 is a police officer.
### Response:
forall x: P(x) -> Q(x). forall x: Q(x) or ~R(x). forall x: P(x) or R(x). forall x: S(x). Therefore forall x: Q(x) and S(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? There is at least one x_12 for which x_12 will not work. For all x_12, x_12 is a police officer if and only if x_12 is a janitor. For all x_12, x_12 is a police officer if and only if x_12 will work. | For all x_12, if x_12 were a police officer, then x_12 is a janitor. For all x_12, if x_12 were a police officer, then x_12 is working. For all x_12, if x_12 were a janitor, then x_12 is a police officer. For all x_12, if x_12 works, then x_12 is a police officer. There is at least one x_12 for which x_12 is not a police officer. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? There is at least one x_12 for which x_12 will not work. For all x_12, x_12 is a police officer if and only if x_12 is a janitor. For all x_12, x_12 is a police officer if and only if x_12 will work.
### Response:
For all x_12, if x_12 were a police officer, then x_12 is a janitor. For all x_12, if x_12 were a police officer, then x_12 is working. For all x_12, if x_12 were a janitor, then x_12 is a police officer. For all x_12, if x_12 works, then x_12 is a police officer. There is at least one x_12 for which x_12 is not a police officer.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_20, x_20 is a carpenter and x_20 is a scientist. For all x_9, x_9 is fast. | For all x_20, x_20 is a carpenter. For all x_20, x_20 is a scientist. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_20, x_20 is a carpenter and x_20 is a scientist. For all x_9, x_9 is fast.
### Response:
For all x_20, x_20 is a carpenter. For all x_20, x_20 is a scientist.
| null | null |
Consider the following premises. There is at least one x_1 for which x_1 is not excited. For all x_1, x_1 will write a letter if and only if x_1 is excited. For all x_1, x_1 will write a letter or x_1 will not drive a car. Can we infer the following from them? For all x_1, x_1 will drive a car. | From the fact that for all x_1, x_1 will write a letter or x_1 will not drive a car we can infer that for all x_1, if x_1 drives a car, then x_1 is writing a letter. From the fact that for all x_1, x_1 will write a letter if and only if x_1 is excited, and that there is at least one x_1 for which x_1 is not excited we can infer that there is at least one x_1 for which x_1 will not write a letter. Finally, from the fact that for all x_1, if x_1 drives a car, then x_1 is writing a letter, and that there is at least one x_1 for which x_1 will not write a letter we can infer that there is at least one x_1 for which x_1 will not drive a car, which contradicts that For all x_1, x_1 will drive a car. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. There is at least one x_1 for which x_1 is not excited. For all x_1, x_1 will write a letter if and only if x_1 is excited. For all x_1, x_1 will write a letter or x_1 will not drive a car. Can we infer the following from them? For all x_1, x_1 will drive a car.
### Response:
From the fact that for all x_1, x_1 will write a letter or x_1 will not drive a car we can infer that for all x_1, if x_1 drives a car, then x_1 is writing a letter. From the fact that for all x_1, x_1 will write a letter if and only if x_1 is excited, and that there is at least one x_1 for which x_1 is not excited we can infer that there is at least one x_1 for which x_1 will not write a letter. Finally, from the fact that for all x_1, if x_1 drives a car, then x_1 is writing a letter, and that there is at least one x_1 for which x_1 will not write a letter we can infer that there is at least one x_1 for which x_1 will not drive a car, which contradicts that For all x_1, x_1 will drive a car. Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_5: P_5(x_5). forall x_5: W_4(x_5) -> Q(x_5). forall x_5: ~P_5(x_5). | Nothing can be inferred from these premises. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_5: P_5(x_5). forall x_5: W_4(x_5) -> Q(x_5). forall x_5: ~P_5(x_5).
### Response:
Nothing can be inferred from these premises.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~q. s -> r_3. r_3 -> q. s or ~q. | ~r_3 can be inferred via the modus tollens rule. s -> q can be inferred via the transitivity rule. q -> s can be inferred via the rewriting rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~q. s -> r_3. r_3 -> q. s or ~q.
### Response:
~r_3 can be inferred via the modus tollens rule. s -> q can be inferred via the transitivity rule. q -> s can be inferred via the rewriting rule.
| null | null |
Consider the following premises. forall x_6: P_6(x_6). forall x_6: ~P_6(x_6). forall x_6: U_2(x_6). Can we infer exists x_6: ~Q(x_6) from them? | The premises are contradictory and we can infer anything from them. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_6: P_6(x_6). forall x_6: ~P_6(x_6). forall x_6: U_2(x_6). Can we infer exists x_6: ~Q(x_6) from them?
### Response:
The premises are contradictory and we can infer anything from them. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: If it is late, then Jennifer is taking a plane. If it is cold, then Elisabeth is sleeping. If Elisabeth sleeps, then it is late. If Jennifer takes a plane, then it is cold. | If it is late, then it is cold can be inferred via the transitivity rule. If it is cold, then it is late can be inferred via the transitivity rule. If Elisabeth sleeps, then Jennifer is taking a plane can be inferred via the transitivity rule. If Jennifer takes a plane, then Elisabeth is sleeping can be inferred via the transitivity rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: If it is late, then Jennifer is taking a plane. If it is cold, then Elisabeth is sleeping. If Elisabeth sleeps, then it is late. If Jennifer takes a plane, then it is cold.
### Response:
If it is late, then it is cold can be inferred via the transitivity rule. If it is cold, then it is late can be inferred via the transitivity rule. If Elisabeth sleeps, then Jennifer is taking a plane can be inferred via the transitivity rule. If Jennifer takes a plane, then Elisabeth is sleeping can be inferred via the transitivity rule.
| null | null |
Consider the following premises. Sarah is not excited. Sarah is excited if and only if Richard is cooking. Richard is cooking or it is late. Can we infer the following from them? If we can, name the inference rule being used: Patricia is sad or it is late. | From the fact that Sarah is excited if and only if Richard is cooking, and that Sarah is not excited we can infer that Richard is not cooking via biconditional elimination. From the fact that Richard is not cooking we can infer that Richard is not cooking or Patricia is sad via addition. Finally, from the fact that Richard is not cooking or Patricia is sad, and that Richard is cooking or it is late we can infer that Patricia is sad or it is late via resolution. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. Sarah is not excited. Sarah is excited if and only if Richard is cooking. Richard is cooking or it is late. Can we infer the following from them? If we can, name the inference rule being used: Patricia is sad or it is late.
### Response:
From the fact that Sarah is excited if and only if Richard is cooking, and that Sarah is not excited we can infer that Richard is not cooking via biconditional elimination. From the fact that Richard is not cooking we can infer that Richard is not cooking or Patricia is sad via addition. Finally, from the fact that Richard is not cooking or Patricia is sad, and that Richard is cooking or it is late we can infer that Patricia is sad or it is late via resolution. Therefore, the answer is yes.
| null | null |
Translate the following inference to logic notation: If Joseph were a nurse, then Joseph is making tea. If Joseph makes tea, then Joseph is a nurse. If Joseph makes tea, then Joseph is a nurse. Therefore if Joseph makes tea, then Joseph is a nurse. | p -> q. q -> p. q -> p. Therefore q -> p. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: If Joseph were a nurse, then Joseph is making tea. If Joseph makes tea, then Joseph is a nurse. If Joseph makes tea, then Joseph is a nurse. Therefore if Joseph makes tea, then Joseph is a nurse.
### Response:
p -> q. q -> p. q -> p. Therefore q -> p.
| null | null |
Translate the following inference to logic notation: There is at least one x for which x is an astronaut if and only if x is an electrician. For all x, x is an electrician. For all x, x is a writer. Therefore there is at least one x for which x is a writer and x is an astronaut. | exists x: P(x) <-> Q(x). forall x: Q(x). forall x: R(x). Therefore exists x: R(x) and P(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: There is at least one x for which x is an astronaut if and only if x is an electrician. For all x, x is an electrician. For all x, x is a writer. Therefore there is at least one x for which x is a writer and x is an astronaut.
### Response:
exists x: P(x) <-> Q(x). forall x: Q(x). forall x: R(x). Therefore exists x: R(x) and P(x).
| null | null |
Consider the following premises. p -> u_4. p <-> r_2. u_4 -> r_2. Can we infer p -> r_2 from them? | From p <-> r_2 we can infer r_2 -> p. Finally, from p -> u_4, u_4 -> r_2 we can infer p -> r_2. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. p -> u_4. p <-> r_2. u_4 -> r_2. Can we infer p -> r_2 from them?
### Response:
From p <-> r_2 we can infer r_2 -> p. Finally, from p -> u_4, u_4 -> r_2 we can infer p -> r_2. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: Karen is sleeping if and only if James is a taxi driver. For all x_19, x_19 is not sad. Jessica is sad or it is cloudy. If James were a taxi driver, then it is not cloudy. | If Karen sleeps, then James is a taxi driver can be inferred via the biconditional elimination rule. If James were a taxi driver, then Karen is sleeping can be inferred via the biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: Karen is sleeping if and only if James is a taxi driver. For all x_19, x_19 is not sad. Jessica is sad or it is cloudy. If James were a taxi driver, then it is not cloudy.
### Response:
If Karen sleeps, then James is a taxi driver can be inferred via the biconditional elimination rule. If James were a taxi driver, then Karen is sleeping can be inferred via the biconditional elimination rule.
| null | null |
Translate the following inference to logic notation: For all x, if y_3 were a musician, then y_3 is writing a letter. For all x, y_3 will write a letter or y_3 is an office clerk. For all x, if y_3 were an office clerk, then y_3 is a musician. There is at least one x for which if y_3 writes a letter, then y_3 is a musician. Therefore there is at least one x for which y_3 is a musician. | forall x: P(x) -> Q(x). forall x: Q(x) or R(x). forall x: R(x) -> P(x). exists x: Q(x) -> P(x). Therefore exists x: P(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, if y_3 were a musician, then y_3 is writing a letter. For all x, y_3 will write a letter or y_3 is an office clerk. For all x, if y_3 were an office clerk, then y_3 is a musician. There is at least one x for which if y_3 writes a letter, then y_3 is a musician. Therefore there is at least one x for which y_3 is a musician.
### Response:
forall x: P(x) -> Q(x). forall x: Q(x) or R(x). forall x: R(x) -> P(x). exists x: Q(x) -> P(x). Therefore exists x: P(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~Q_2(i_4). t -> S_7(i_4). forall x_7: Q_2(x_7) <-> S_7(x_7). | forall x_7: Q_2(x_7) -> S_7(x_7) can be inferred via the universal biconditional elimination rule. forall x_7: S_7(x_7) -> Q_2(x_7) can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: ~Q_2(i_4). t -> S_7(i_4). forall x_7: Q_2(x_7) <-> S_7(x_7).
### Response:
forall x_7: Q_2(x_7) -> S_7(x_7) can be inferred via the universal biconditional elimination rule. forall x_7: S_7(x_7) -> Q_2(x_7) can be inferred via the universal biconditional elimination rule.
| null | null |
Consider the following premises. For all x_9, x_9 will play a game. For all x_8, if x_8 plays a game, then x_8 is listening to a song. If it is late, then James is listening to a song. Can we infer the following from them? If it is late, then James is listening to a song. | That is one of the premises. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_9, x_9 will play a game. For all x_8, if x_8 plays a game, then x_8 is listening to a song. If it is late, then James is listening to a song. Can we infer the following from them? If it is late, then James is listening to a song.
### Response:
That is one of the premises. Therefore, the answer is yes.
| null | null |
Consider the following premises. forall x_5: P_5(x_5) -> Q_5(x_5). forall x_5: Q_5(x_5) -> R_5(x_5). forall x_5: ~P_2(x_5). forall x_5: P_2(x_5) or P_5(x_5). Can we infer exists x_5: ~R_5(x_5) from them? | From forall x_5: P_5(x_5) -> Q_5(x_5), forall x_5: Q_5(x_5) -> R_5(x_5) we can infer forall x_5: P_5(x_5) -> R_5(x_5). From forall x_5: P_2(x_5) or P_5(x_5), forall x_5: ~P_2(x_5) we can infer forall x_5: P_5(x_5). Finally, from forall x_5: P_5(x_5) -> R_5(x_5), forall x_5: P_5(x_5) we can infer forall x_5: R_5(x_5), which contradicts exists x_5: ~R_5(x_5). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_5: P_5(x_5) -> Q_5(x_5). forall x_5: Q_5(x_5) -> R_5(x_5). forall x_5: ~P_2(x_5). forall x_5: P_2(x_5) or P_5(x_5). Can we infer exists x_5: ~R_5(x_5) from them?
### Response:
From forall x_5: P_5(x_5) -> Q_5(x_5), forall x_5: Q_5(x_5) -> R_5(x_5) we can infer forall x_5: P_5(x_5) -> R_5(x_5). From forall x_5: P_2(x_5) or P_5(x_5), forall x_5: ~P_2(x_5) we can infer forall x_5: P_5(x_5). Finally, from forall x_5: P_5(x_5) -> R_5(x_5), forall x_5: P_5(x_5) we can infer forall x_5: R_5(x_5), which contradicts exists x_5: ~R_5(x_5). Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_7: R(x_7). forall x_7: S_4(x_7) -> ~R(x_7). S_4(a_5) <-> q. | S_4(a_5) -> q can be inferred via the biconditional elimination rule. q -> S_4(a_5) can be inferred via the biconditional elimination rule. forall x_7: ~S_4(x_7) can be inferred via the universal modus tollens rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_7: R(x_7). forall x_7: S_4(x_7) -> ~R(x_7). S_4(a_5) <-> q.
### Response:
S_4(a_5) -> q can be inferred via the biconditional elimination rule. q -> S_4(a_5) can be inferred via the biconditional elimination rule. forall x_7: ~S_4(x_7) can be inferred via the universal modus tollens rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: w. p <-> u. ~p. | w and ~p can be inferred via the conjunction rule. ~p and w can be inferred via the conjunction rule. p -> u can be inferred via the biconditional elimination rule. u -> p can be inferred via the biconditional elimination rule. ~u can be inferred via the biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: w. p <-> u. ~p.
### Response:
w and ~p can be inferred via the conjunction rule. ~p and w can be inferred via the conjunction rule. p -> u can be inferred via the biconditional elimination rule. u -> p can be inferred via the biconditional elimination rule. ~u can be inferred via the biconditional elimination rule.
| null | null |
Consider the following premises. ~q_6. forall y_3: U_2(y_3) <-> U_1(y_3). U_1(a_12) <-> p_6. p_6 -> q_6. Can we infer U_2(a_12) from them? If possible, name the inference rules being used at each step. | From forall y_3: U_2(y_3) <-> U_1(y_3) we can infer U_2(a_12) <-> U_1(a_12) via universal instantiation. From p_6 -> q_6, ~q_6 we can infer ~p_6 via modus tollens. From U_1(a_12) <-> p_6, ~p_6 we can infer ~U_1(a_12) via biconditional elimination. Finally, from U_2(a_12) <-> U_1(a_12), ~U_1(a_12) we can infer ~U_2(a_12) via biconditional elimination, which contradicts U_2(a_12). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. ~q_6. forall y_3: U_2(y_3) <-> U_1(y_3). U_1(a_12) <-> p_6. p_6 -> q_6. Can we infer U_2(a_12) from them? If possible, name the inference rules being used at each step.
### Response:
From forall y_3: U_2(y_3) <-> U_1(y_3) we can infer U_2(a_12) <-> U_1(a_12) via universal instantiation. From p_6 -> q_6, ~q_6 we can infer ~p_6 via modus tollens. From U_1(a_12) <-> p_6, ~p_6 we can infer ~U_1(a_12) via biconditional elimination. Finally, from U_2(a_12) <-> U_1(a_12), ~U_1(a_12) we can infer ~U_2(a_12) via biconditional elimination, which contradicts U_2(a_12). Therefore, the answer is no.
| null | null |
Translate the following inference to logic notation: For all x, x_8 is a janitor. For all x, x_8 will drink water and x_8 will sleep. Therefore for all x, x_8 is a janitor or x_8 will make tea. | forall x: P(x). forall x: Q(x) and R(x). Therefore forall x: P(x) or S(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, x_8 is a janitor. For all x, x_8 will drink water and x_8 will sleep. Therefore for all x, x_8 is a janitor or x_8 will make tea.
### Response:
forall x: P(x). forall x: Q(x) and R(x). Therefore forall x: P(x) or S(x).
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_1: R(x_1) <-> T(x_1). forall x_1: R(x_1) <-> U(x_1). | forall x_1: R(x_1) -> T(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: R(x_1) -> U(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: T(x_1) -> R(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: U(x_1) -> R(x_1) can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall x_1: R(x_1) <-> T(x_1). forall x_1: R(x_1) <-> U(x_1).
### Response:
forall x_1: R(x_1) -> T(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: R(x_1) -> U(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: T(x_1) -> R(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: U(x_1) -> R(x_1) can be inferred via the universal biconditional elimination rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: P_9(d) <-> r_3. forall x_9: P_9(x_9) <-> Q_9(x_9). forall x_9: ~Q_9(x_9). | P_9(d) -> r_3 can be inferred via the biconditional elimination rule. r_3 -> P_9(d) can be inferred via the biconditional elimination rule. forall x_9: P_9(x_9) -> Q_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_9(x_9) -> P_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~P_9(x_9) can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: P_9(d) <-> r_3. forall x_9: P_9(x_9) <-> Q_9(x_9). forall x_9: ~Q_9(x_9).
### Response:
P_9(d) -> r_3 can be inferred via the biconditional elimination rule. r_3 -> P_9(d) can be inferred via the biconditional elimination rule. forall x_9: P_9(x_9) -> Q_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_9(x_9) -> P_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~P_9(x_9) can be inferred via the universal biconditional elimination rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? q -> p. p -> q. p. | q. q <-> p. p <-> q. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? q -> p. p -> q. p.
### Response:
q. q <-> p. p <-> q.
| null | null |
Consider the following premises. ~t. t <-> q_4. Can we infer q_4 from them? | From t <-> q_4 we can infer q_4 -> t. Finally, from q_4 -> t, ~t we can infer ~q_4, which contradicts q_4. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. ~t. t <-> q_4. Can we infer q_4 from them?
### Response:
From t <-> q_4 we can infer q_4 -> t. Finally, from q_4 -> t, ~t we can infer ~q_4, which contradicts q_4. Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_7, x_7 is a lawyer or x_7 will not drive a car. There is at least one x_7 for which if x_7 were a lawyer, then x_7 is running. For all x_7, x_7 is not a lawyer. For all x_7, if x_7 goes running, then x_7 is driving a car. | There is at least one x_7 for which if x_7 were a lawyer, then x_7 is driving a car. For all x_7, if x_7 drives a car, then x_7 is a lawyer. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_7, x_7 is a lawyer or x_7 will not drive a car. There is at least one x_7 for which if x_7 were a lawyer, then x_7 is running. For all x_7, x_7 is not a lawyer. For all x_7, if x_7 goes running, then x_7 is driving a car.
### Response:
There is at least one x_7 for which if x_7 were a lawyer, then x_7 is driving a car. For all x_7, if x_7 drives a car, then x_7 is a lawyer.
| null | null |
Consider the following premises. p_4 <-> p_9. p_9 <-> u. Can we infer u -> p_4 from them? If possible, name the inference rules being used at each step. | From p_9 <-> u we can infer u -> p_9 via biconditional elimination. From p_4 <-> p_9 we can infer p_9 -> p_4 via biconditional elimination. Finally, from u -> p_9, p_9 -> p_4 we can infer u -> p_4 via transitivity. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. p_4 <-> p_9. p_9 <-> u. Can we infer u -> p_4 from them? If possible, name the inference rules being used at each step.
### Response:
From p_9 <-> u we can infer u -> p_9 via biconditional elimination. From p_4 <-> p_9 we can infer p_9 -> p_4 via biconditional elimination. Finally, from u -> p_9, p_9 -> p_4 we can infer u -> p_4 via transitivity. Therefore, the answer is yes.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: p_4 -> w_4. p_4 or ~w_4. | w_4 -> p_4 can be inferred via the rewriting rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: p_4 -> w_4. p_4 or ~w_4.
### Response:
w_4 -> p_4 can be inferred via the rewriting rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_11, x_11 is a politician if and only if x_11 will sleep. For all x_11, if x_11 sleeps, then x_11 is a lawyer. There is at least one x_11 for which x_11 is not a lawyer. | There is at least one x_11 for which x_11 will not sleep. For all x_11, if x_11 were a politician, then x_11 is sleeping. For all x_11, if x_11 sleeps, then x_11 is a politician. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_11, x_11 is a politician if and only if x_11 will sleep. For all x_11, if x_11 sleeps, then x_11 is a lawyer. There is at least one x_11 for which x_11 is not a lawyer.
### Response:
There is at least one x_11 for which x_11 will not sleep. For all x_11, if x_11 were a politician, then x_11 is sleeping. For all x_11, if x_11 sleeps, then x_11 is a politician.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all y, y will not play squash. For all y, if y drinks water, then y is playing squash. | For all y, y will not drink water can be inferred via the universal modus tollens rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all y, y will not play squash. For all y, if y drinks water, then y is playing squash.
### Response:
For all y, y will not drink water can be inferred via the universal modus tollens rule.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: S_3(a_7) -> p. forall z: ~S_3(z). p -> S_3(a_7). | Nothing can be inferred from these premises. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: S_3(a_7) -> p. forall z: ~S_3(z). p -> S_3(a_7).
### Response:
Nothing can be inferred from these premises.
| null | null |
Translate the following inference to logic notation: For all x, x_15 is a writer. For all x, if x_15 were a writer, then x_15 is an astronaut. There is at least one x for which x_15 is an astronaut if and only if x_15 is rich. Therefore there is at least one x for which x_15 is rich or x_15 is happy. | forall x: P(x). forall x: P(x) -> Q(x). exists x: Q(x) <-> R(x). Therefore exists x: R(x) or S(x). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: For all x, x_15 is a writer. For all x, if x_15 were a writer, then x_15 is an astronaut. There is at least one x for which x_15 is an astronaut if and only if x_15 is rich. Therefore there is at least one x for which x_15 is rich or x_15 is happy.
### Response:
forall x: P(x). forall x: P(x) -> Q(x). exists x: Q(x) <-> R(x). Therefore exists x: R(x) or S(x).
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Consider the following premises. There is at least one z_4 for which z_4 is a mechanic if and only if z_4 is a lawyer. For all z_4, if z_4 were a lawyer, then z_4 is a janitor. For all z_4, z_4 is a mechanic. Can we infer the following from them? There is at least one z_4 for which z_4 is a janitor. | From the fact that there is at least one z_4 for which z_4 is a mechanic if and only if z_4 is a lawyer we can infer that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a lawyer. From the fact that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a lawyer, and that for all z_4, if z_4 were a lawyer, then z_4 is a janitor we can infer that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a janitor. Finally, from the fact that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a janitor, and that for all z_4, z_4 is a mechanic we can infer that there is at least one z_4 for which z_4 is a janitor. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. There is at least one z_4 for which z_4 is a mechanic if and only if z_4 is a lawyer. For all z_4, if z_4 were a lawyer, then z_4 is a janitor. For all z_4, z_4 is a mechanic. Can we infer the following from them? There is at least one z_4 for which z_4 is a janitor.
### Response:
From the fact that there is at least one z_4 for which z_4 is a mechanic if and only if z_4 is a lawyer we can infer that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a lawyer. From the fact that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a lawyer, and that for all z_4, if z_4 were a lawyer, then z_4 is a janitor we can infer that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a janitor. Finally, from the fact that there is at least one z_4 for which if z_4 were a mechanic, then z_4 is a janitor, and that for all z_4, z_4 is a mechanic we can infer that there is at least one z_4 for which z_4 is a janitor. Therefore, the answer is yes.
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Consider the following premises. r_3 -> p_5. p_5 <-> r_3. ~r_3. Can we infer ~p_5 from them? | From p_5 <-> r_3 we can infer p_5 -> r_3. From r_3 -> p_5, p_5 -> r_3 we can infer r_3 <-> p_5. Finally, from r_3 <-> p_5, ~r_3 we can infer ~p_5. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. r_3 -> p_5. p_5 <-> r_3. ~r_3. Can we infer ~p_5 from them?
### Response:
From p_5 <-> r_3 we can infer p_5 -> r_3. From r_3 -> p_5, p_5 -> r_3 we can infer r_3 <-> p_5. Finally, from r_3 <-> p_5, ~r_3 we can infer ~p_5. Therefore, the answer is yes.
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Consider the following premises. For all x_5, x_5 will work or x_5 will not play tennis. For all x_5, if x_5 climbs a mountain, then x_5 is making tea. For all x_5, if x_5 makes tea, then x_5 is playing tennis. Can we infer the following from them? For all x_5, if x_5 climbs a mountain, then x_5 is making tea. | That is one of the premises. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_5, x_5 will work or x_5 will not play tennis. For all x_5, if x_5 climbs a mountain, then x_5 is making tea. For all x_5, if x_5 makes tea, then x_5 is playing tennis. Can we infer the following from them? For all x_5, if x_5 climbs a mountain, then x_5 is making tea.
### Response:
That is one of the premises. Therefore, the answer is yes.
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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_1: R(x_1) <-> P(x_1). forall x_1: P(x_1) <-> R(x_1). | forall x_1: R(x_1) -> P(x_1). forall x_1: P(x_1) -> R(x_1). forall x_1: P(x_1) -> R(x_1). forall x_1: R(x_1) -> P(x_1). | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_1: R(x_1) <-> P(x_1). forall x_1: P(x_1) <-> R(x_1).
### Response:
forall x_1: R(x_1) -> P(x_1). forall x_1: P(x_1) -> R(x_1). forall x_1: P(x_1) -> R(x_1). forall x_1: R(x_1) -> P(x_1).
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Consider the following premises. q_4 -> p. p -> q_4. t <-> p. Can we infer t from them? If possible, name the inference rules being used at each step. | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. q_4 -> p. p -> q_4. t <-> p. Can we infer t from them? If possible, name the inference rules being used at each step.
### Response:
We cannot infer that from the premises. Therefore the answer is no.
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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_8, x_8 is fast if and only if x_8 is a politician. For all x_8, x_8 is fast or x_8 will take a plane. For all x_8, x_8 is not a politician. For all x_8, x_8 will play a game. | For all x_8, if x_8 were fast, then x_8 is a politician. For all x_8, if x_8 were a politician, then x_8 is fast. For all x_8, x_8 is not fast. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_8, x_8 is fast if and only if x_8 is a politician. For all x_8, x_8 is fast or x_8 will take a plane. For all x_8, x_8 is not a politician. For all x_8, x_8 will play a game.
### Response:
For all x_8, if x_8 were fast, then x_8 is a politician. For all x_8, if x_8 were a politician, then x_8 is fast. For all x_8, x_8 is not fast.
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Consider the following premises. For all x_6, x_6 is an astronaut or x_6 is a cashier. For all x_6, x_6 is not an astronaut. For all x_6, if x_6 were a cashier, then x_6 is a writer. For all x_6, if x_6 were a writer, then x_6 is cooking. Can we infer the following from them? There is at least one x_6 for which x_6 will not cook. | From the fact that for all x_6, x_6 is an astronaut or x_6 is a cashier, and that for all x_6, x_6 is not an astronaut we can infer that for all x_6, x_6 is a cashier. From the fact that for all x_6, if x_6 were a cashier, then x_6 is a writer, and that for all x_6, x_6 is a cashier we can infer that for all x_6, x_6 is a writer. Finally, from the fact that for all x_6, if x_6 were a writer, then x_6 is cooking, and that for all x_6, x_6 is a writer we can infer that for all x_6, x_6 will cook, which contradicts that There is at least one x_6 for which x_6 will not cook. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_6, x_6 is an astronaut or x_6 is a cashier. For all x_6, x_6 is not an astronaut. For all x_6, if x_6 were a cashier, then x_6 is a writer. For all x_6, if x_6 were a writer, then x_6 is cooking. Can we infer the following from them? There is at least one x_6 for which x_6 will not cook.
### Response:
From the fact that for all x_6, x_6 is an astronaut or x_6 is a cashier, and that for all x_6, x_6 is not an astronaut we can infer that for all x_6, x_6 is a cashier. From the fact that for all x_6, if x_6 were a cashier, then x_6 is a writer, and that for all x_6, x_6 is a cashier we can infer that for all x_6, x_6 is a writer. Finally, from the fact that for all x_6, if x_6 were a writer, then x_6 is cooking, and that for all x_6, x_6 is a writer we can infer that for all x_6, x_6 will cook, which contradicts that There is at least one x_6 for which x_6 will not cook. Therefore, the answer is no.
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Consider the following premises. forall x_5: P_2(x_5) or P(x_5). exists x_5: W_4(x_5) or ~R_4(x_5). forall x_5: ~P_2(x_5) or R_4(x_5). forall x_5: P(x_5) -> W_4(x_5). Can we infer exists x_5: W_4(x_5) from them? If possible, name the inference rules being used at each step. | From exists x_5: W_4(x_5) or ~R_4(x_5) we can infer exists x_5: R_4(x_5) -> W_4(x_5) via existential rewriting. From forall x_5: P_2(x_5) or P(x_5), forall x_5: ~P_2(x_5) or R_4(x_5) we can infer forall x_5: P(x_5) or R_4(x_5) via universal resolution. Finally, from forall x_5: P(x_5) -> W_4(x_5), exists x_5: R_4(x_5) -> W_4(x_5), forall x_5: P(x_5) or R_4(x_5) we can infer exists x_5: W_4(x_5) via existential disjunction elimination. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_5: P_2(x_5) or P(x_5). exists x_5: W_4(x_5) or ~R_4(x_5). forall x_5: ~P_2(x_5) or R_4(x_5). forall x_5: P(x_5) -> W_4(x_5). Can we infer exists x_5: W_4(x_5) from them? If possible, name the inference rules being used at each step.
### Response:
From exists x_5: W_4(x_5) or ~R_4(x_5) we can infer exists x_5: R_4(x_5) -> W_4(x_5) via existential rewriting. From forall x_5: P_2(x_5) or P(x_5), forall x_5: ~P_2(x_5) or R_4(x_5) we can infer forall x_5: P(x_5) or R_4(x_5) via universal resolution. Finally, from forall x_5: P(x_5) -> W_4(x_5), exists x_5: R_4(x_5) -> W_4(x_5), forall x_5: P(x_5) or R_4(x_5) we can infer exists x_5: W_4(x_5) via existential disjunction elimination. Therefore, the answer is yes.
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Translate the following inference to logic notation: If Jessica were bored, then Linda is a mechanic. Jessica is bored and it is raining. Michael is cooking. Therefore Linda is a mechanic and Michael is cooking. | p -> q. p and r. s. Therefore q and s. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: If Jessica were bored, then Linda is a mechanic. Jessica is bored and it is raining. Michael is cooking. Therefore Linda is a mechanic and Michael is cooking.
### Response:
p -> q. p and r. s. Therefore q and s.
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Consider the following premises. exists y_4: T(y_4) -> W(y_4). forall y_4: T(y_4) or ~W(y_4). Can we infer forall y_4: ~(W(y_4) -> T(y_4)) from them? If possible, name the inference rules being used at each step. | That contradicts the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. exists y_4: T(y_4) -> W(y_4). forall y_4: T(y_4) or ~W(y_4). Can we infer forall y_4: ~(W(y_4) -> T(y_4)) from them? If possible, name the inference rules being used at each step.
### Response:
That contradicts the premises. Therefore the answer is no.
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Consider the following premises. q_2 <-> q_22. t -> ~u. u. ~q_2. Can we infer ~q_22 -> ~t from them? | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. q_2 <-> q_22. t -> ~u. u. ~q_2. Can we infer ~q_22 -> ~t from them?
### Response:
We cannot infer that from the premises. Therefore the answer is no.
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Consider the following premises. For all x_8, x_8 will not write a letter. For all z_1, if z_1 plays a game, then z_1 is sleeping. For all x_8, x_8 will write a letter or x_8 will play a game. Can we infer the following from them? John is not sleeping. | From the fact that for all x_8, x_8 will write a letter or x_8 will play a game, and that for all x_8, x_8 will not write a letter we can infer that for all x_8, x_8 will play a game. From the fact that for all x_8, x_8 will play a game we can infer that John is playing a game. From the fact that for all z_1, if z_1 plays a game, then z_1 is sleeping we can infer that if John plays a game, then John is sleeping. Finally, from the fact that if John plays a game, then John is sleeping, and that John is playing a game we can infer that John is sleeping, which contradicts that John is not sleeping. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_8, x_8 will not write a letter. For all z_1, if z_1 plays a game, then z_1 is sleeping. For all x_8, x_8 will write a letter or x_8 will play a game. Can we infer the following from them? John is not sleeping.
### Response:
From the fact that for all x_8, x_8 will write a letter or x_8 will play a game, and that for all x_8, x_8 will not write a letter we can infer that for all x_8, x_8 will play a game. From the fact that for all x_8, x_8 will play a game we can infer that John is playing a game. From the fact that for all z_1, if z_1 plays a game, then z_1 is sleeping we can infer that if John plays a game, then John is sleeping. Finally, from the fact that if John plays a game, then John is sleeping, and that John is playing a game we can infer that John is sleeping, which contradicts that John is not sleeping. Therefore, the answer is no.
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Consider the following premises. John is a scientist. If it snows, then John is not a scientist. If Sarah reads a book, then Richard is not listening to a song. It is snowing or Richard is listening to a song. Can we infer the following from them? If we can, name the inference rule being used: Barbara is a writer. | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. John is a scientist. If it snows, then John is not a scientist. If Sarah reads a book, then Richard is not listening to a song. It is snowing or Richard is listening to a song. Can we infer the following from them? If we can, name the inference rule being used: Barbara is a writer.
### Response:
We cannot infer that from the premises. Therefore the answer is no.
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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all x_5, x_5 will play squash if and only if x_5 is an astronaut. For all x_5, if x_5 makes tea, then x_5 is bored. For all x_5, if x_5 were an astronaut, then x_5 is making tea. For all x_5, x_5 is not bored. | For all x_5, x_5 will not make tea can be inferred via the universal modus tollens rule. For all x_5, if x_5 were an astronaut, then x_5 is bored can be inferred via the universal transitivity rule. For all x_5, if x_5 plays squash, then x_5 is an astronaut can be inferred via the universal biconditional elimination rule. For all x_5, if x_5 were an astronaut, then x_5 is playing squash can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all x_5, x_5 will play squash if and only if x_5 is an astronaut. For all x_5, if x_5 makes tea, then x_5 is bored. For all x_5, if x_5 were an astronaut, then x_5 is making tea. For all x_5, x_5 is not bored.
### Response:
For all x_5, x_5 will not make tea can be inferred via the universal modus tollens rule. For all x_5, if x_5 were an astronaut, then x_5 is bored can be inferred via the universal transitivity rule. For all x_5, if x_5 plays squash, then x_5 is an astronaut can be inferred via the universal biconditional elimination rule. For all x_5, if x_5 were an astronaut, then x_5 is playing squash can be inferred via the universal biconditional elimination rule.
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Consider the following premises. forall x_9: Q_1(x_9) -> P_9(x_9). exists x_9: W_2(x_9) -> Q_1(x_9). forall x_9: P_9(x_9) <-> Q_9(x_9). forall x_9: W_2(x_9) or Q_9(x_9). Can we infer exists x_9: P_9(x_9) from them? If possible, name the inference rules being used at each step. | From forall x_9: P_9(x_9) <-> Q_9(x_9) we can infer forall x_9: Q_9(x_9) -> P_9(x_9) via universal biconditional elimination. From exists x_9: W_2(x_9) -> Q_1(x_9), forall x_9: Q_1(x_9) -> P_9(x_9) we can infer exists x_9: W_2(x_9) -> P_9(x_9) via existential transitivity. Finally, from exists x_9: W_2(x_9) -> P_9(x_9), forall x_9: Q_9(x_9) -> P_9(x_9), forall x_9: W_2(x_9) or Q_9(x_9) we can infer exists x_9: P_9(x_9) via existential disjunction elimination. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_9: Q_1(x_9) -> P_9(x_9). exists x_9: W_2(x_9) -> Q_1(x_9). forall x_9: P_9(x_9) <-> Q_9(x_9). forall x_9: W_2(x_9) or Q_9(x_9). Can we infer exists x_9: P_9(x_9) from them? If possible, name the inference rules being used at each step.
### Response:
From forall x_9: P_9(x_9) <-> Q_9(x_9) we can infer forall x_9: Q_9(x_9) -> P_9(x_9) via universal biconditional elimination. From exists x_9: W_2(x_9) -> Q_1(x_9), forall x_9: Q_1(x_9) -> P_9(x_9) we can infer exists x_9: W_2(x_9) -> P_9(x_9) via existential transitivity. Finally, from exists x_9: W_2(x_9) -> P_9(x_9), forall x_9: Q_9(x_9) -> P_9(x_9), forall x_9: W_2(x_9) or Q_9(x_9) we can infer exists x_9: P_9(x_9) via existential disjunction elimination. Therefore, the answer is yes.
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What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all z, if z makes tea, then z is listening to a song. For all z, z is a mechanic or z will make tea. For all z, z will read a book if and only if z is a mechanic. For all z, if z listens to a song, then z is reading a book. | For all z, if z makes tea, then z is reading a book can be inferred via the universal transitivity rule. For all z, if z reads a book, then z is a mechanic can be inferred via the universal biconditional elimination rule. For all z, if z were a mechanic, then z is reading a book can be inferred via the universal biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: For all z, if z makes tea, then z is listening to a song. For all z, z is a mechanic or z will make tea. For all z, z will read a book if and only if z is a mechanic. For all z, if z listens to a song, then z is reading a book.
### Response:
For all z, if z makes tea, then z is reading a book can be inferred via the universal transitivity rule. For all z, if z reads a book, then z is a mechanic can be inferred via the universal biconditional elimination rule. For all z, if z were a mechanic, then z is reading a book can be inferred via the universal biconditional elimination rule.
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Consider the following premises. For all z, z is an astronaut. Thomas is listening to a song. If it is cold, then Susan is not an astronaut. Can we infer the following from them? If Thomas listens to a song, then it is not cold. | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all z, z is an astronaut. Thomas is listening to a song. If it is cold, then Susan is not an astronaut. Can we infer the following from them? If Thomas listens to a song, then it is not cold.
### Response:
We cannot infer that from the premises. Therefore the answer is no.
| null | null |
Consider the following premises. u <-> s. ~t_1. u <-> u_3. r <-> t_1. r <-> u_3. Can we infer s from them? If possible, name the inference rules being used at each step. | From r <-> t_1, ~t_1 we can infer ~r via biconditional elimination. From r <-> u_3, ~r we can infer ~u_3 via biconditional elimination. From u <-> u_3, ~u_3 we can infer ~u via biconditional elimination. Finally, from u <-> s, ~u we can infer ~s via biconditional elimination, which contradicts s. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. u <-> s. ~t_1. u <-> u_3. r <-> t_1. r <-> u_3. Can we infer s from them? If possible, name the inference rules being used at each step.
### Response:
From r <-> t_1, ~t_1 we can infer ~r via biconditional elimination. From r <-> u_3, ~r we can infer ~u_3 via biconditional elimination. From u <-> u_3, ~u_3 we can infer ~u via biconditional elimination. Finally, from u <-> s, ~u we can infer ~s via biconditional elimination, which contradicts s. Therefore, the answer is no.
| null | null |
Translate the following inference to logic notation: If it rains, then Elisabeth is rich. Elisabeth is rich if and only if it is cold. If David were an astronaut, then Elisabeth is rich. David is an astronaut or it is raining. Therefore it is cold. | p -> q. q <-> r. s -> q. s or p. Therefore r. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: If it rains, then Elisabeth is rich. Elisabeth is rich if and only if it is cold. If David were an astronaut, then Elisabeth is rich. David is an astronaut or it is raining. Therefore it is cold.
### Response:
p -> q. q <-> r. s -> q. s or p. Therefore r.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? t or t_1. q_4 <-> t. t_1 -> q_4. | q_4 -> t. t -> q_4. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? t or t_1. q_4 <-> t. t_1 -> q_4.
### Response:
q_4 -> t. t -> q_4.
| null | null |
Consider the following premises. For all x_5, x_5 will work or x_5 will not make tea. For all x_5, if x_5 were a writer, then x_5 is cooking. There is at least one x_5 for which if x_5 works, then x_5 is a writer. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_5 for which if x_5 makes tea, then x_5 is cooking. | From the fact that there is at least one x_5 for which if x_5 works, then x_5 is a writer, and that for all x_5, if x_5 were a writer, then x_5 is cooking we can infer that there is at least one x_5 for which if x_5 works, then x_5 is cooking via existential transitivity. From the fact that for all x_5, x_5 will work or x_5 will not make tea we can infer that for all x_5, if x_5 makes tea, then x_5 is working via universal rewriting. Finally, from the fact that for all x_5, if x_5 makes tea, then x_5 is working, and that there is at least one x_5 for which if x_5 works, then x_5 is cooking we can infer that there is at least one x_5 for which if x_5 makes tea, then x_5 is cooking via existential transitivity. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. For all x_5, x_5 will work or x_5 will not make tea. For all x_5, if x_5 were a writer, then x_5 is cooking. There is at least one x_5 for which if x_5 works, then x_5 is a writer. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_5 for which if x_5 makes tea, then x_5 is cooking.
### Response:
From the fact that there is at least one x_5 for which if x_5 works, then x_5 is a writer, and that for all x_5, if x_5 were a writer, then x_5 is cooking we can infer that there is at least one x_5 for which if x_5 works, then x_5 is cooking via existential transitivity. From the fact that for all x_5, x_5 will work or x_5 will not make tea we can infer that for all x_5, if x_5 makes tea, then x_5 is working via universal rewriting. Finally, from the fact that for all x_5, if x_5 makes tea, then x_5 is working, and that there is at least one x_5 for which if x_5 works, then x_5 is cooking we can infer that there is at least one x_5 for which if x_5 makes tea, then x_5 is cooking via existential transitivity. Therefore, the answer is yes.
| null | null |
Consider the following premises. forall x_5: S(x_5) -> ~U(x_5). forall x_5: P_5(x_5) or U(x_5). exists x_5: ~P_5(x_5). Can we infer forall x_5: S(x_5) from them? | From forall x_5: P_5(x_5) or U(x_5), exists x_5: ~P_5(x_5) we can infer exists x_5: U(x_5). Finally, from forall x_5: S(x_5) -> ~U(x_5), exists x_5: U(x_5) we can infer exists x_5: ~S(x_5), which contradicts forall x_5: S(x_5). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. forall x_5: S(x_5) -> ~U(x_5). forall x_5: P_5(x_5) or U(x_5). exists x_5: ~P_5(x_5). Can we infer forall x_5: S(x_5) from them?
### Response:
From forall x_5: P_5(x_5) or U(x_5), exists x_5: ~P_5(x_5) we can infer exists x_5: U(x_5). Finally, from forall x_5: S(x_5) -> ~U(x_5), exists x_5: U(x_5) we can infer exists x_5: ~S(x_5), which contradicts forall x_5: S(x_5). Therefore, the answer is no.
| null | null |
Consider the following premises. p_6 or ~r_2. p_6 -> q_6. q_6 -> r_2. Can we infer ~(r_2 <-> p_6) from them? | From p_6 -> q_6, q_6 -> r_2 we can infer p_6 -> r_2. From p_6 or ~r_2 we can infer r_2 -> p_6. Finally, from r_2 -> p_6, p_6 -> r_2 we can infer r_2 <-> p_6, which contradicts ~(r_2 <-> p_6). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. p_6 or ~r_2. p_6 -> q_6. q_6 -> r_2. Can we infer ~(r_2 <-> p_6) from them?
### Response:
From p_6 -> q_6, q_6 -> r_2 we can infer p_6 -> r_2. From p_6 or ~r_2 we can infer r_2 -> p_6. Finally, from r_2 -> p_6, p_6 -> r_2 we can infer r_2 <-> p_6, which contradicts ~(r_2 <-> p_6). Therefore, the answer is no.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? There is at least one x_11 for which if x_11 works, then x_11 is writing a letter. For all x_11, if x_11 writes a letter, then x_11 is an artist. For all x_11, if x_11 were a carpenter, then x_11 is a lawyer. For all x_11, if x_11 were a lawyer, then x_11 is working. For all x_11, if x_11 were a nurse, then x_11 is a lawyer. For all x_11, x_11 is a nurse or x_11 is a carpenter. | For all x_11, if x_11 were a carpenter, then x_11 is working. For all x_11, if x_11 were a nurse, then x_11 is working. There is at least one x_11 for which if x_11 works, then x_11 is an artist. There is at least one x_11 for which if x_11 were a lawyer, then x_11 is writing a letter. For all x_11, x_11 is a lawyer. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? There is at least one x_11 for which if x_11 works, then x_11 is writing a letter. For all x_11, if x_11 writes a letter, then x_11 is an artist. For all x_11, if x_11 were a carpenter, then x_11 is a lawyer. For all x_11, if x_11 were a lawyer, then x_11 is working. For all x_11, if x_11 were a nurse, then x_11 is a lawyer. For all x_11, x_11 is a nurse or x_11 is a carpenter.
### Response:
For all x_11, if x_11 were a carpenter, then x_11 is working. For all x_11, if x_11 were a nurse, then x_11 is working. There is at least one x_11 for which if x_11 works, then x_11 is an artist. There is at least one x_11 for which if x_11 were a lawyer, then x_11 is writing a letter. For all x_11, x_11 is a lawyer.
| null | null |
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall z_1: Q(z_1) -> U(z_1). exists z_1: ~U(z_1). forall z_1: U(z_1) <-> Q(z_1). | exists z_1: ~Q(z_1) can be inferred via the existential modus tollens rule. forall z_1: U(z_1) -> Q(z_1) can be inferred via the universal biconditional elimination rule. forall z_1: Q(z_1) -> U(z_1) can be inferred via the universal biconditional elimination rule. exists z_1: ~Q(z_1) can be inferred via the existential biconditional elimination rule. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Name the inference rule being used: forall z_1: Q(z_1) -> U(z_1). exists z_1: ~U(z_1). forall z_1: U(z_1) <-> Q(z_1).
### Response:
exists z_1: ~Q(z_1) can be inferred via the existential modus tollens rule. forall z_1: U(z_1) -> Q(z_1) can be inferred via the universal biconditional elimination rule. forall z_1: Q(z_1) -> U(z_1) can be inferred via the universal biconditional elimination rule. exists z_1: ~Q(z_1) can be inferred via the existential biconditional elimination rule.
| null | null |
Consider the following premises. ~u. forall x_10: R_10(x_10) -> S_10(x_10). S_10(a_10) -> u. Can we infer R_10(a_10) from them? | From forall x_10: R_10(x_10) -> S_10(x_10) we can infer R_10(a_10) -> S_10(a_10). From S_10(a_10) -> u, ~u we can infer ~S_10(a_10). Finally, from R_10(a_10) -> S_10(a_10), ~S_10(a_10) we can infer ~R_10(a_10), which contradicts R_10(a_10). Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. ~u. forall x_10: R_10(x_10) -> S_10(x_10). S_10(a_10) -> u. Can we infer R_10(a_10) from them?
### Response:
From forall x_10: R_10(x_10) -> S_10(x_10) we can infer R_10(a_10) -> S_10(a_10). From S_10(a_10) -> u, ~u we can infer ~S_10(a_10). Finally, from R_10(a_10) -> S_10(a_10), ~S_10(a_10) we can infer ~R_10(a_10), which contradicts R_10(a_10). Therefore, the answer is no.
| null | null |
Consider the following premises. If it is overcast, then it is raining. It is overcast if and only if it is cloudy. It is not cloudy. If it rains, then it is overcast. Can we infer the following from them? If we can, name the inference rule being used: It is raining. | From the fact that if it is overcast, then it is raining, and that if it rains, then it is overcast we can infer that it is overcast if and only if it is raining via biconditional introduction. From the fact that it is overcast if and only if it is cloudy, and that it is not cloudy we can infer that it is not overcast via biconditional elimination. Finally, from the fact that it is overcast if and only if it is raining, and that it is not overcast we can infer that it is not raining via biconditional elimination, which contradicts that It is raining. Therefore, the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. If it is overcast, then it is raining. It is overcast if and only if it is cloudy. It is not cloudy. If it rains, then it is overcast. Can we infer the following from them? If we can, name the inference rule being used: It is raining.
### Response:
From the fact that if it is overcast, then it is raining, and that if it rains, then it is overcast we can infer that it is overcast if and only if it is raining via biconditional introduction. From the fact that it is overcast if and only if it is cloudy, and that it is not cloudy we can infer that it is not overcast via biconditional elimination. Finally, from the fact that it is overcast if and only if it is raining, and that it is not overcast we can infer that it is not raining via biconditional elimination, which contradicts that It is raining. Therefore, the answer is no.
| null | null |
Consider the following premises. u_1 or p_12. u_1 -> q_12. p_12 <-> q_12. Can we infer ~w_2 from them? | We cannot infer that from the premises. Therefore the answer is no. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. u_1 or p_12. u_1 -> q_12. p_12 <-> q_12. Can we infer ~w_2 from them?
### Response:
We cannot infer that from the premises. Therefore the answer is no.
| null | null |
Consider the following premises. James is an astronaut. If Elisabeth cooks, then James is not an astronaut. If Elisabeth doesn't cook, then Sarah is a server. David is writing a letter. Can we infer the following from them? If we can, name the inference rule being used: Sarah is a server and David is writing a letter. | From the fact that if Elisabeth cooks, then James is not an astronaut, and that James is an astronaut we can infer that Elisabeth is not cooking via modus tollens. From the fact that if Elisabeth doesn't cook, then Sarah is a server, and that Elisabeth is not cooking we can infer that Sarah is a server via modus ponens. Finally, from the fact that Sarah is a server, and that David is writing a letter we can infer that Sarah is a server and David is writing a letter via conjunction. Therefore, the answer is yes. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Consider the following premises. James is an astronaut. If Elisabeth cooks, then James is not an astronaut. If Elisabeth doesn't cook, then Sarah is a server. David is writing a letter. Can we infer the following from them? If we can, name the inference rule being used: Sarah is a server and David is writing a letter.
### Response:
From the fact that if Elisabeth cooks, then James is not an astronaut, and that James is an astronaut we can infer that Elisabeth is not cooking via modus tollens. From the fact that if Elisabeth doesn't cook, then Sarah is a server, and that Elisabeth is not cooking we can infer that Sarah is a server via modus ponens. Finally, from the fact that Sarah is a server, and that David is writing a letter we can infer that Sarah is a server and David is writing a letter via conjunction. Therefore, the answer is yes.
| null | null |
Translate the following inference to logic notation: If it rains, then Thomas is a cashier. If Thomas were a cashier, then it is snowing. It is snowing if and only if it is raining. Therefore it is snowing if and only if it is raining. | p -> q. q -> r. r <-> p. Therefore r <-> p. | LogicInference Dataset e | ### Instruction:
Answer the following.
### Input:
Translate the following inference to logic notation: If it rains, then Thomas is a cashier. If Thomas were a cashier, then it is snowing. It is snowing if and only if it is raining. Therefore it is snowing if and only if it is raining.
### Response:
p -> q. q -> r. r <-> p. Therefore r <-> p.
| null | null |
End of preview. Expand
in Dataset Viewer.
This is just a concatenation of several other data sets mostly converted to Alpaca style prompts to help give a good logic data set to settle some models or fine tune.
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