Datasets:

ibivibiv

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variety-logic-training

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Translate the following inference to logic notation: For all x, x_14 is a carpenter or x_14 will make tea. David is a server. Thomas is not a carpenter. Therefore Thomas is making tea.

forall x: P(x) or Q(x). p. ~P(a). Therefore Q(a).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, x_14 is a carpenter or x_14 will make tea. David is a server. Thomas is not a carpenter. Therefore Thomas is making tea. ### Response: forall x: P(x) or Q(x). p. ~P(a). Therefore Q(a).

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_9: ~Q_7(x_9). forall x_9: Q_9(x_9) -> R_9(x_9). forall x_9: P_9(x_9) -> Q_9(x_9). forall x_9: R_9(x_9) <-> Q_7(x_9).

forall x_9: P_9(x_9) -> R_9(x_9) can be inferred via the universal transitivity rule. forall x_9: R_9(x_9) -> Q_7(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_7(x_9) -> R_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~R_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: forall x_9: ~Q_7(x_9). forall x_9: Q_9(x_9) -> R_9(x_9). forall x_9: P_9(x_9) -> Q_9(x_9). forall x_9: R_9(x_9) <-> Q_7(x_9). ### Response: forall x_9: P_9(x_9) -> R_9(x_9) can be inferred via the universal transitivity rule. forall x_9: R_9(x_9) -> Q_7(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_7(x_9) -> R_9(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~R_9(x_9) can be inferred via the universal biconditional elimination rule.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? q <-> S(a_4). forall x_1: S(x_1).

q -> S(a_4). S(a_4) -> 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 <-> S(a_4). forall x_1: S(x_1). ### Response: q -> S(a_4). S(a_4) -> q.

Consider the following premises. For all x_5, x_5 will cook or x_5 will not go running. For all x_5, if x_5 were a carpenter, then x_5 is running. For all x_5, if x_5 cooks, then x_5 is a carpenter. Can we infer the following from them? For all x_5, x_5 will go running if and only if x_5 will cook.

From the fact that for all x_5, if x_5 cooks, then x_5 is a carpenter, and that for all x_5, if x_5 were a carpenter, then x_5 is running we can infer that for all x_5, if x_5 cooks, then x_5 is running. From the fact that for all x_5, x_5 will cook or x_5 will not go running we can infer that for all x_5, if x_5 goes running, then x_5 is cooking. Finally, from the fact that for all x_5, if x_5 goes running, then x_5 is cooking, and that for all x_5, if x_5 cooks, then x_5 is running we can infer that for all x_5, x_5 will go running if and only if x_5 will cook. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. For all x_5, x_5 will cook or x_5 will not go running. For all x_5, if x_5 were a carpenter, then x_5 is running. For all x_5, if x_5 cooks, then x_5 is a carpenter. Can we infer the following from them? For all x_5, x_5 will go running if and only if x_5 will cook. ### Response: From the fact that for all x_5, if x_5 cooks, then x_5 is a carpenter, and that for all x_5, if x_5 were a carpenter, then x_5 is running we can infer that for all x_5, if x_5 cooks, then x_5 is running. From the fact that for all x_5, x_5 will cook or x_5 will not go running we can infer that for all x_5, if x_5 goes running, then x_5 is cooking. Finally, from the fact that for all x_5, if x_5 goes running, then x_5 is cooking, and that for all x_5, if x_5 cooks, then x_5 is running we can infer that for all x_5, x_5 will go running if and only if x_5 will cook. Therefore, the answer is yes.

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_8, x_8 is not a poet. For all x_8, x_8 is a lawyer. For all x_8, if x_8 were a lawyer, then x_8 is a poet.

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)? Name the inference rule being used: For all x_8, x_8 is not a poet. For all x_8, x_8 is a lawyer. For all x_8, if x_8 were a lawyer, then x_8 is a poet. ### Response: Since the premises are contradictory, we can infer anything from them.

Consider the following premises. forall x_16: P_16(x_16) <-> Q_16(x_16). forall x_16: P_16(x_16) or ~Q_11(x_16). Q_11(a_5). Can we infer ~r_4 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. forall x_16: P_16(x_16) <-> Q_16(x_16). forall x_16: P_16(x_16) or ~Q_11(x_16). Q_11(a_5). Can we infer ~r_4 from them? ### Response: We cannot infer that from the premises. Therefore the answer is no.

Consider the following premises. p -> p_5. p_2 or q_5. ~p_2 or p. p_5 or ~q_5. Can we infer p_5 from them?

From p_5 or ~q_5 we can infer q_5 -> p_5. From p_2 or q_5, ~p_2 or p we can infer q_5 or p. Finally, from q_5 -> p_5, p -> p_5, q_5 or p we can infer p_5. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. p -> p_5. p_2 or q_5. ~p_2 or p. p_5 or ~q_5. Can we infer p_5 from them? ### Response: From p_5 or ~q_5 we can infer q_5 -> p_5. From p_2 or q_5, ~p_2 or p we can infer q_5 or p. Finally, from q_5 -> p_5, p -> p_5, q_5 or p we can infer p_5. Therefore, the answer is yes.

Translate the following inference to logic notation: There is at least one x for which if y plays squash, then y is cooking. For all x, y will cook if and only if y will play squash. Therefore there is at least one x for which if y plays squash, then y is cooking.

exists x: P(x) -> Q(x). forall x: Q(x) <-> P(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 if y plays squash, then y is cooking. For all x, y will cook if and only if y will play squash. Therefore there is at least one x for which if y plays squash, then y is cooking. ### Response: exists x: P(x) -> Q(x). forall x: Q(x) <-> P(x). Therefore exists x: P(x) -> Q(x).

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_10, if x_10 drives a car, then x_10 is a janitor. There is at least one x_10 for which if x_10 were bored, then x_10 is a janitor. For all x_10, if x_10 were a janitor, then x_10 is bored. For all x_10, x_10 is not bored.

For all x_10, x_10 is not a janitor. For all x_10, if x_10 drives a car, then x_10 is bored. There is at least one x_10 for which x_10 is bored if and only if x_10 is a janitor. There is at least one x_10 for which x_10 is a janitor if and only if x_10 is bored.

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_10, if x_10 drives a car, then x_10 is a janitor. There is at least one x_10 for which if x_10 were bored, then x_10 is a janitor. For all x_10, if x_10 were a janitor, then x_10 is bored. For all x_10, x_10 is not bored. ### Response: For all x_10, x_10 is not a janitor. For all x_10, if x_10 drives a car, then x_10 is bored. There is at least one x_10 for which x_10 is bored if and only if x_10 is a janitor. There is at least one x_10 for which x_10 is a janitor if and only if x_10 is bored.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_11: R_11(x_11) <-> S_11(x_11). ~R_11(a_11). w_3.

~R_11(a_11) and w_3. w_3 and ~R_11(a_11). forall x_11: R_11(x_11) -> S_11(x_11). forall x_11: S_11(x_11) -> R_11(x_11).

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_11: R_11(x_11) <-> S_11(x_11). ~R_11(a_11). w_3. ### Response: ~R_11(a_11) and w_3. w_3 and ~R_11(a_11). forall x_11: R_11(x_11) -> S_11(x_11). forall x_11: S_11(x_11) -> R_11(x_11).

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_14: P_14(x_14) <-> Q_14(x_14). forall x_14: Q(x_14) <-> Q_6(x_14). exists x_14: ~P_14(x_14). forall x_14: Q_14(x_14) <-> Q_6(x_14).

forall x_14: P_14(x_14) -> Q_14(x_14). forall x_14: Q(x_14) -> Q_6(x_14). forall x_14: Q_14(x_14) -> Q_6(x_14). forall x_14: Q_14(x_14) -> P_14(x_14). forall x_14: Q_6(x_14) -> Q(x_14). forall x_14: Q_6(x_14) -> Q_14(x_14). exists x_14: ~Q_14(x_14).

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_14: P_14(x_14) <-> Q_14(x_14). forall x_14: Q(x_14) <-> Q_6(x_14). exists x_14: ~P_14(x_14). forall x_14: Q_14(x_14) <-> Q_6(x_14). ### Response: forall x_14: P_14(x_14) -> Q_14(x_14). forall x_14: Q(x_14) -> Q_6(x_14). forall x_14: Q_14(x_14) -> Q_6(x_14). forall x_14: Q_14(x_14) -> P_14(x_14). forall x_14: Q_6(x_14) -> Q(x_14). forall x_14: Q_6(x_14) -> Q_14(x_14). exists x_14: ~Q_14(x_14).

Consider the following premises. For all x_11, x_11 is a lawyer if and only if x_11 is excited. For all x_11, x_11 is a lawyer if and only if x_11 is excited. Can we infer the following from them? The fact that John were a lawyer does not imply that John is excited.

That contradicts the premises. Therefore the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. For all x_11, x_11 is a lawyer if and only if x_11 is excited. For all x_11, x_11 is a lawyer if and only if x_11 is excited. Can we infer the following from them? The fact that John were a lawyer does not imply that John is excited. ### Response: That contradicts the premises. Therefore the answer is no.

Consider the following premises. For all y_3, y_3 is an astronaut or y_3 will play squash. There is at least one y_3 for which if y_3 plays squash, then y_3 is driving a car. For all y_3, y_3 will drive a car if and only if y_3 is an astronaut. Can we infer the following from them? For all y_3, y_3 will not drive a car.

From the fact that for all y_3, y_3 will drive a car if and only if y_3 is an astronaut we can infer that for all y_3, if y_3 were an astronaut, then y_3 is driving a car. Finally, from the fact that for all y_3, if y_3 were an astronaut, then y_3 is driving a car, that there is at least one y_3 for which if y_3 plays squash, then y_3 is driving a car, and that for all y_3, y_3 is an astronaut or y_3 will play squash we can infer that there is at least one y_3 for which y_3 will drive a car, which contradicts that For all y_3, y_3 will not drive a car. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. For all y_3, y_3 is an astronaut or y_3 will play squash. There is at least one y_3 for which if y_3 plays squash, then y_3 is driving a car. For all y_3, y_3 will drive a car if and only if y_3 is an astronaut. Can we infer the following from them? For all y_3, y_3 will not drive a car. ### Response: From the fact that for all y_3, y_3 will drive a car if and only if y_3 is an astronaut we can infer that for all y_3, if y_3 were an astronaut, then y_3 is driving a car. Finally, from the fact that for all y_3, if y_3 were an astronaut, then y_3 is driving a car, that there is at least one y_3 for which if y_3 plays squash, then y_3 is driving a car, and that for all y_3, y_3 is an astronaut or y_3 will play squash we can infer that there is at least one y_3 for which y_3 will drive a car, which contradicts that For all y_3, y_3 will not drive a car. Therefore, the answer is no.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? u or ~q. q_1 -> u. s. q or q_1.

q -> u.

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 or ~q. q_1 -> u. s. q or q_1. ### Response: q -> u.

Consider the following premises. It is not windy. It is windy if and only if David is playing squash. It is overcast if and only if Patricia is listening to a song. For all x_23, x_23 will not listen to a song. Can we infer the following from them? It is overcast.

From the fact that for all x_23, x_23 will not listen to a song we can infer that Patricia is not listening to a song. Finally, from the fact that it is overcast if and only if Patricia is listening to a song, and that Patricia is not listening to a song we can infer that it is not overcast, which contradicts that It is overcast. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. It is not windy. It is windy if and only if David is playing squash. It is overcast if and only if Patricia is listening to a song. For all x_23, x_23 will not listen to a song. Can we infer the following from them? It is overcast. ### Response: From the fact that for all x_23, x_23 will not listen to a song we can infer that Patricia is not listening to a song. Finally, from the fact that it is overcast if and only if Patricia is listening to a song, and that Patricia is not listening to a song we can infer that it is not overcast, which contradicts that It is overcast. Therefore, the answer is no.

Consider the following premises. forall x_18: P_18(x_18). P_18(a_18) -> r_2. Can we infer r_2 or q from them? If possible, name the inference rules being used at each step.

From forall x_18: P_18(x_18) we can infer P_18(a_18) via universal instantiation. From P_18(a_18) -> r_2, P_18(a_18) we can infer r_2 via modus ponens. Finally, from r_2 we can infer r_2 or q via addition. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_18: P_18(x_18). P_18(a_18) -> r_2. Can we infer r_2 or q from them? If possible, name the inference rules being used at each step. ### Response: From forall x_18: P_18(x_18) we can infer P_18(a_18) via universal instantiation. From P_18(a_18) -> r_2, P_18(a_18) we can infer r_2 via modus ponens. Finally, from r_2 we can infer r_2 or q via addition. Therefore, the answer is yes.

Consider the following premises. w_2 -> w. ~w. w <-> w_2. Can we infer ~w_2 from them?

From w <-> w_2 we can infer w -> w_2. From w -> w_2, w_2 -> w we can infer w <-> w_2. From w <-> w_2 we can infer w_2 -> w. Finally, from w_2 -> w, ~w we can infer ~w_2. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. w_2 -> w. ~w. w <-> w_2. Can we infer ~w_2 from them? ### Response: From w <-> w_2 we can infer w -> w_2. From w -> w_2, w_2 -> w we can infer w <-> w_2. From w <-> w_2 we can infer w_2 -> w. Finally, from w_2 -> w, ~w we can infer ~w_2. Therefore, the answer is yes.

Consider the following premises. w_3 -> s. u. w_3. Can we infer ~s from them? If possible, name the inference rules being used at each step.

From w_3, u we can infer w_3 and u via conjunction. From w_3 and u we can infer w_3 via simplification. Finally, from w_3 -> s, w_3 we can infer s via modus ponens, which contradicts ~s. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. w_3 -> s. u. w_3. Can we infer ~s from them? If possible, name the inference rules being used at each step. ### Response: From w_3, u we can infer w_3 and u via conjunction. From w_3 and u we can infer w_3 via simplification. Finally, from w_3 -> s, w_3 we can infer s via modus ponens, which contradicts ~s. Therefore, the answer is no.

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: S(z) -> U(z). exists z: U(z) <-> S(z).

exists z: U(z) -> S(z) can be inferred via the existential biconditional elimination rule. exists z: S(z) -> U(z) 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: S(z) -> U(z). exists z: U(z) <-> S(z). ### Response: exists z: U(z) -> S(z) can be inferred via the existential biconditional elimination rule. exists z: S(z) -> U(z) can be inferred via the existential biconditional elimination rule.

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_2: P(z_2) or W_1(z_2). ~q or r. ~P(a_9) or q. ~W_1(a_9) or w.

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 z_2: P(z_2) or W_1(z_2). ~q or r. ~P(a_9) or q. ~W_1(a_9) or w. ### Response: Nothing can be inferred from these premises.

Consider the following premises. forall x_8: P_8(x_8) or P(x_8). forall x_8: ~Q_8(x_8). forall x_8: P_8(x_8) <-> Q_8(x_8). Can we infer forall x_8: P(x_8) or P_3(x_8) from them? If possible, name the inference rules being used at each step.

From forall x_8: P_8(x_8) <-> Q_8(x_8), forall x_8: ~Q_8(x_8) we can infer forall x_8: ~P_8(x_8) via universal biconditional elimination. From forall x_8: P_8(x_8) or P(x_8), forall x_8: ~P_8(x_8) we can infer forall x_8: P(x_8) via universal disjunctive syllogism. Finally, from forall x_8: P(x_8) we can infer forall x_8: P(x_8) or P_3(x_8) via universal addition. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_8: P_8(x_8) or P(x_8). forall x_8: ~Q_8(x_8). forall x_8: P_8(x_8) <-> Q_8(x_8). Can we infer forall x_8: P(x_8) or P_3(x_8) from them? If possible, name the inference rules being used at each step. ### Response: From forall x_8: P_8(x_8) <-> Q_8(x_8), forall x_8: ~Q_8(x_8) we can infer forall x_8: ~P_8(x_8) via universal biconditional elimination. From forall x_8: P_8(x_8) or P(x_8), forall x_8: ~P_8(x_8) we can infer forall x_8: P(x_8) via universal disjunctive syllogism. Finally, from forall x_8: P(x_8) we can infer forall x_8: P(x_8) or P_3(x_8) via universal addition. Therefore, the answer is yes.

Consider the following premises. forall x_7: ~P_7(x_7). forall x_7: Q_7(x_7) or Q_2(x_7). forall x_7: W(x_7). forall x_7: P_7(x_7) <-> Q_7(x_7). Can we infer forall x_7: W(x_7) and Q_2(x_7) from them?

From forall x_7: P_7(x_7) <-> Q_7(x_7), forall x_7: ~P_7(x_7) we can infer forall x_7: ~Q_7(x_7). From forall x_7: Q_7(x_7) or Q_2(x_7), forall x_7: ~Q_7(x_7) we can infer forall x_7: Q_2(x_7). Finally, from forall x_7: W(x_7), forall x_7: Q_2(x_7) we can infer forall x_7: W(x_7) and Q_2(x_7). Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_7: ~P_7(x_7). forall x_7: Q_7(x_7) or Q_2(x_7). forall x_7: W(x_7). forall x_7: P_7(x_7) <-> Q_7(x_7). Can we infer forall x_7: W(x_7) and Q_2(x_7) from them? ### Response: From forall x_7: P_7(x_7) <-> Q_7(x_7), forall x_7: ~P_7(x_7) we can infer forall x_7: ~Q_7(x_7). From forall x_7: Q_7(x_7) or Q_2(x_7), forall x_7: ~Q_7(x_7) we can infer forall x_7: Q_2(x_7). Finally, from forall x_7: W(x_7), forall x_7: Q_2(x_7) we can infer forall x_7: W(x_7) and Q_2(x_7). Therefore, the answer is yes.

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_2 or ~u_4. u_4 and p.

u_4 can be inferred via the simplification rule. p can be inferred via the simplification rule. u_4 -> p_2 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_2 or ~u_4. u_4 and p. ### Response: u_4 can be inferred via the simplification rule. p can be inferred via the simplification rule. u_4 -> p_2 can be inferred via the rewriting rule.

Consider the following premises. For all x_1, if x_1 were bored, then x_1 is not a politician. For all x_1, x_1 is rich. For all x_1, if x_1 were rich, then x_1 is a politician. Can we infer the following from them? There is at least one x_1 for which x_1 is bored.

From the fact that for all x_1, if x_1 were rich, then x_1 is a politician, and that for all x_1, x_1 is rich we can infer that for all x_1, x_1 is a politician. Finally, from the fact that for all x_1, if x_1 were bored, then x_1 is not a politician, and that for all x_1, x_1 is a politician we can infer that for all x_1, x_1 is not bored, which contradicts that There is at least one x_1 for which x_1 is bored. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. For all x_1, if x_1 were bored, then x_1 is not a politician. For all x_1, x_1 is rich. For all x_1, if x_1 were rich, then x_1 is a politician. Can we infer the following from them? There is at least one x_1 for which x_1 is bored. ### Response: From the fact that for all x_1, if x_1 were rich, then x_1 is a politician, and that for all x_1, x_1 is rich we can infer that for all x_1, x_1 is a politician. Finally, from the fact that for all x_1, if x_1 were bored, then x_1 is not a politician, and that for all x_1, x_1 is a politician we can infer that for all x_1, x_1 is not bored, which contradicts that There is at least one x_1 for which x_1 is bored. Therefore, the answer is no.

Translate the following inference to logic notation: For all x, y is a carpenter or y will sleep. For all x, y is a carpenter if and only if y is a lawyer. For all x, if y were a lawyer, then y is happy. There is at least one x for which if y sleeps, then y is happy. Therefore there is at least one x for which y is happy.

forall x: P(x) or Q(x). forall x: P(x) <-> R(x). forall x: R(x) -> S(x). exists x: Q(x) -> S(x). Therefore exists x: S(x).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, y is a carpenter or y will sleep. For all x, y is a carpenter if and only if y is a lawyer. For all x, if y were a lawyer, then y is happy. There is at least one x for which if y sleeps, then y is happy. Therefore there is at least one x for which y is happy. ### Response: forall x: P(x) or Q(x). forall x: P(x) <-> R(x). forall x: R(x) -> S(x). exists x: Q(x) -> S(x). Therefore exists x: S(x).

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_3 for which x_3 will read a book. For all x_3, if x_3 were excited, then x_3 is poor. For all x_3, x_3 is excited if and only if x_3 will read a book.

For all x_3, if x_3 were excited, then x_3 is reading a book. For all x_3, if x_3 reads a book, then x_3 is excited.

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_3 for which x_3 will read a book. For all x_3, if x_3 were excited, then x_3 is poor. For all x_3, x_3 is excited if and only if x_3 will read a book. ### Response: For all x_3, if x_3 were excited, then x_3 is reading a book. For all x_3, if x_3 reads a book, then x_3 is excited.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_11: T(x_11) -> ~P_11(x_11). exists x_11: P_9(x_11) or Q_11(x_11). forall x_11: P_11(x_11) or ~Q_11(x_11). forall x_11: P_9(x_11) -> P_11(x_11).

forall x_11: Q_11(x_11) -> P_11(x_11).

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_11: T(x_11) -> ~P_11(x_11). exists x_11: P_9(x_11) or Q_11(x_11). forall x_11: P_11(x_11) or ~Q_11(x_11). forall x_11: P_9(x_11) -> P_11(x_11). ### Response: forall x_11: Q_11(x_11) -> P_11(x_11).

Consider the following premises. forall x_14: R_2(x_14) <-> S_14(x_14). S_14(a_14) -> ~t_1. R_2(a_14) -> w. w -> R_2(a_14). t_1. Can we infer w from them?

From forall x_14: R_2(x_14) <-> S_14(x_14) we can infer R_2(a_14) <-> S_14(a_14). From S_14(a_14) -> ~t_1, t_1 we can infer ~S_14(a_14). From R_2(a_14) -> w, w -> R_2(a_14) we can infer R_2(a_14) <-> w. From R_2(a_14) <-> S_14(a_14), ~S_14(a_14) we can infer ~R_2(a_14). Finally, from R_2(a_14) <-> w, ~R_2(a_14) we can infer ~w, which contradicts w. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_14: R_2(x_14) <-> S_14(x_14). S_14(a_14) -> ~t_1. R_2(a_14) -> w. w -> R_2(a_14). t_1. Can we infer w from them? ### Response: From forall x_14: R_2(x_14) <-> S_14(x_14) we can infer R_2(a_14) <-> S_14(a_14). From S_14(a_14) -> ~t_1, t_1 we can infer ~S_14(a_14). From R_2(a_14) -> w, w -> R_2(a_14) we can infer R_2(a_14) <-> w. From R_2(a_14) <-> S_14(a_14), ~S_14(a_14) we can infer ~R_2(a_14). Finally, from R_2(a_14) <-> w, ~R_2(a_14) we can infer ~w, which contradicts w. Therefore, the answer is no.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Sarah is writing a letter or Robert is not a doctor. Patricia is cooking or Sarah is not writing a letter.

If Robert were a doctor, then Sarah is writing a letter. If Sarah writes a letter, then Patricia 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)? Sarah is writing a letter or Robert is not a doctor. Patricia is cooking or Sarah is not writing a letter. ### Response: If Robert were a doctor, then Sarah is writing a letter. If Sarah writes a letter, then Patricia is cooking.

Consider the following premises. w. w -> p. p -> s. ~s or t_4. Can we infer t_4 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. w. w -> p. p -> s. ~s or t_4. Can we infer t_4 from them? ### Response: We cannot infer that from the premises. Therefore the answer is no.

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_7, x_7 is not a cashier. There is at least one x_7 for which x_7 will drive a car if and only if x_7 is a cashier. For all x_7, x_7 will drive a car or x_7 will drink water.

There is at least one x_7 for which if x_7 drives a car, then x_7 is a cashier can be inferred via the existential biconditional elimination rule. There is at least one x_7 for which if x_7 were a cashier, then x_7 is driving a car can be inferred via the existential biconditional elimination rule. There is at least one x_7 for which x_7 will not drive a car 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: For all x_7, x_7 is not a cashier. There is at least one x_7 for which x_7 will drive a car if and only if x_7 is a cashier. For all x_7, x_7 will drive a car or x_7 will drink water. ### Response: There is at least one x_7 for which if x_7 drives a car, then x_7 is a cashier can be inferred via the existential biconditional elimination rule. There is at least one x_7 for which if x_7 were a cashier, then x_7 is driving a car can be inferred via the existential biconditional elimination rule. There is at least one x_7 for which x_7 will not drive a car can be inferred via the existential biconditional elimination rule.

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: Barbara is excited if and only if Elisabeth is drinking water. For all z, z will not drink water.

If Barbara were excited, then Elisabeth is drinking water can be inferred via the biconditional elimination rule. If Elisabeth drinks water, then Barbara is excited 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: Barbara is excited if and only if Elisabeth is drinking water. For all z, z will not drink water. ### Response: If Barbara were excited, then Elisabeth is drinking water can be inferred via the biconditional elimination rule. If Elisabeth drinks water, then Barbara is excited can be inferred via the biconditional elimination rule.

Consider the following premises. forall z: U_1(z) -> R(z). forall z: R(z) -> U_1(z). forall z: U(z) -> U_1(z). exists z: U(z) or R(z). Can we infer forall z: ~U_1(z) from them? If possible, name the inference rules being used at each step.

From forall z: R(z) -> U_1(z), forall z: U_1(z) -> R(z) we can infer forall z: R(z) <-> U_1(z) via universal biconditional introduction. From forall z: R(z) <-> U_1(z) we can infer forall z: R(z) -> U_1(z) via universal biconditional elimination. Finally, from forall z: U(z) -> U_1(z), forall z: R(z) -> U_1(z), exists z: U(z) or R(z) we can infer exists z: U_1(z) via existential disjunction elimination, which contradicts forall z: ~U_1(z). Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall z: U_1(z) -> R(z). forall z: R(z) -> U_1(z). forall z: U(z) -> U_1(z). exists z: U(z) or R(z). Can we infer forall z: ~U_1(z) from them? If possible, name the inference rules being used at each step. ### Response: From forall z: R(z) -> U_1(z), forall z: U_1(z) -> R(z) we can infer forall z: R(z) <-> U_1(z) via universal biconditional introduction. From forall z: R(z) <-> U_1(z) we can infer forall z: R(z) -> U_1(z) via universal biconditional elimination. Finally, from forall z: U(z) -> U_1(z), forall z: R(z) -> U_1(z), exists z: U(z) or R(z) we can infer exists z: U_1(z) via existential disjunction elimination, which contradicts forall z: ~U_1(z). Therefore, the answer is no.

Consider the following premises. exists x_6: Q_6(x_6) -> S(x_6). forall x_6: P_1(x_6) -> Q_6(x_6). forall x_6: Q_1(x_6) -> P_1(x_6). forall x_6: S(x_6) -> Q_1(x_6). Can we infer exists x_6: P_1(x_6) <-> S(x_6) from them?

From forall x_6: P_1(x_6) -> Q_6(x_6), exists x_6: Q_6(x_6) -> S(x_6) we can infer exists x_6: P_1(x_6) -> S(x_6). From forall x_6: S(x_6) -> Q_1(x_6), forall x_6: Q_1(x_6) -> P_1(x_6) we can infer forall x_6: S(x_6) -> P_1(x_6). Finally, from exists x_6: P_1(x_6) -> S(x_6), forall x_6: S(x_6) -> P_1(x_6) we can infer exists x_6: P_1(x_6) <-> S(x_6). Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. exists x_6: Q_6(x_6) -> S(x_6). forall x_6: P_1(x_6) -> Q_6(x_6). forall x_6: Q_1(x_6) -> P_1(x_6). forall x_6: S(x_6) -> Q_1(x_6). Can we infer exists x_6: P_1(x_6) <-> S(x_6) from them? ### Response: From forall x_6: P_1(x_6) -> Q_6(x_6), exists x_6: Q_6(x_6) -> S(x_6) we can infer exists x_6: P_1(x_6) -> S(x_6). From forall x_6: S(x_6) -> Q_1(x_6), forall x_6: Q_1(x_6) -> P_1(x_6) we can infer forall x_6: S(x_6) -> P_1(x_6). Finally, from exists x_6: P_1(x_6) -> S(x_6), forall x_6: S(x_6) -> P_1(x_6) we can infer exists x_6: P_1(x_6) <-> S(x_6). Therefore, the answer is yes.

Consider the following premises. forall x_3: P_3(x_3) -> Q_3(x_3). forall x_3: S(x_3) -> ~Q_3(x_3). exists x_3: P_3(x_3). Can we infer exists x_3: ~S(x_3) from them?

From forall x_3: P_3(x_3) -> Q_3(x_3), exists x_3: P_3(x_3) we can infer exists x_3: Q_3(x_3). Finally, from forall x_3: S(x_3) -> ~Q_3(x_3), exists x_3: Q_3(x_3) we can infer exists x_3: ~S(x_3). Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_3: P_3(x_3) -> Q_3(x_3). forall x_3: S(x_3) -> ~Q_3(x_3). exists x_3: P_3(x_3). Can we infer exists x_3: ~S(x_3) from them? ### Response: From forall x_3: P_3(x_3) -> Q_3(x_3), exists x_3: P_3(x_3) we can infer exists x_3: Q_3(x_3). Finally, from forall x_3: S(x_3) -> ~Q_3(x_3), exists x_3: Q_3(x_3) we can infer exists x_3: ~S(x_3). Therefore, the answer is yes.

Consider the following premises. For all x_8, x_8 will play squash or x_8 is a taxi driver. For all x_8, if x_8 were a taxi driver, then x_8 is playing tennis. For all x_8, x_8 is a mechanic and x_8 is fast. For all x_8, if x_8 plays squash, then x_8 is playing tennis. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_8 for which either x_8 is a mechanic or x_8 will play tennis are not true.

From the fact that for all x_8, if x_8 plays squash, then x_8 is playing tennis, that for all x_8, if x_8 were a taxi driver, then x_8 is playing tennis, and that for all x_8, x_8 will play squash or x_8 is a taxi driver we can infer that for all x_8, x_8 will play tennis via universal disjunction elimination. From the fact that for all x_8, x_8 is a mechanic and x_8 is fast we can infer that for all x_8, x_8 is a mechanic via universal simplification. Finally, from the fact that for all x_8, x_8 is a mechanic, and that for all x_8, x_8 will play tennis we can infer that for all x_8, x_8 is a mechanic and x_8 will play tennis via universal conjunction, which contradicts that There is at least one x_8 for which either x_8 is a mechanic or x_8 will play tennis are not true. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. For all x_8, x_8 will play squash or x_8 is a taxi driver. For all x_8, if x_8 were a taxi driver, then x_8 is playing tennis. For all x_8, x_8 is a mechanic and x_8 is fast. For all x_8, if x_8 plays squash, then x_8 is playing tennis. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_8 for which either x_8 is a mechanic or x_8 will play tennis are not true. ### Response: From the fact that for all x_8, if x_8 plays squash, then x_8 is playing tennis, that for all x_8, if x_8 were a taxi driver, then x_8 is playing tennis, and that for all x_8, x_8 will play squash or x_8 is a taxi driver we can infer that for all x_8, x_8 will play tennis via universal disjunction elimination. From the fact that for all x_8, x_8 is a mechanic and x_8 is fast we can infer that for all x_8, x_8 is a mechanic via universal simplification. Finally, from the fact that for all x_8, x_8 is a mechanic, and that for all x_8, x_8 will play tennis we can infer that for all x_8, x_8 is a mechanic and x_8 will play tennis via universal conjunction, which contradicts that There is at least one x_8 for which either x_8 is a mechanic or x_8 will play tennis are not true. Therefore, the answer is no.

Consider the following premises. forall x: ~W(x). exists x: P(x) -> W(x). forall x: P(x) or ~W(x). Can we infer exists x: ~P(x) from them? If possible, name the inference rules being used at each step.

From forall x: P(x) or ~W(x) we can infer forall x: W(x) -> P(x) via universal rewriting. From forall x: W(x) -> P(x), exists x: P(x) -> W(x) we can infer exists x: W(x) <-> P(x) via existential biconditional introduction. Finally, from exists x: W(x) <-> P(x), forall x: ~W(x) we can infer exists x: ~P(x) via existential biconditional elimination. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x: ~W(x). exists x: P(x) -> W(x). forall x: P(x) or ~W(x). Can we infer exists x: ~P(x) from them? If possible, name the inference rules being used at each step. ### Response: From forall x: P(x) or ~W(x) we can infer forall x: W(x) -> P(x) via universal rewriting. From forall x: W(x) -> P(x), exists x: P(x) -> W(x) we can infer exists x: W(x) <-> P(x) via existential biconditional introduction. Finally, from exists x: W(x) <-> P(x), forall x: ~W(x) we can infer exists x: ~P(x) via existential biconditional elimination. Therefore, the answer is yes.

Consider the following premises. R_1(a_9) -> R(a_9). forall z: R(z) <-> R_1(z). Can we infer R(a_9) -> R_1(a_9) from them? If possible, name the inference rules being used at each step.

From forall z: R(z) <-> R_1(z) we can infer R(a_9) <-> R_1(a_9) via universal instantiation. Finally, from R(a_9) <-> R_1(a_9) we can infer R(a_9) -> R_1(a_9) via biconditional elimination. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. R_1(a_9) -> R(a_9). forall z: R(z) <-> R_1(z). Can we infer R(a_9) -> R_1(a_9) from them? If possible, name the inference rules being used at each step. ### Response: From forall z: R(z) <-> R_1(z) we can infer R(a_9) <-> R_1(a_9) via universal instantiation. Finally, from R(a_9) <-> R_1(a_9) we can infer R(a_9) -> R_1(a_9) via biconditional elimination. Therefore, the answer is yes.

Translate the following inference to logic notation: James is not bored or Richard is curious. If Richard were curious, then Jessica is a janitor. If Richard takes a plane, then Jessica is a janitor. James is bored or Richard is taking a plane. Therefore Jessica is a janitor or Karen is drinking water.

~p or q. q -> r. s -> r. p or s. Therefore r or t.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: James is not bored or Richard is curious. If Richard were curious, then Jessica is a janitor. If Richard takes a plane, then Jessica is a janitor. James is bored or Richard is taking a plane. Therefore Jessica is a janitor or Karen is drinking water. ### Response: ~p or q. q -> r. s -> r. p or s. Therefore r or t.

Consider the following premises. forall y: S(y). T(f) <-> q_3. S(f) -> q_3. Can we infer q_3 from them? If possible, name the inference rules being used at each step.

From forall y: S(y) we can infer forall y: S(y) or T(y) via universal addition. From forall y: S(y) or T(y) we can infer S(f) or T(f) via universal instantiation. From T(f) <-> q_3 we can infer T(f) -> q_3 via biconditional elimination. Finally, from S(f) -> q_3, T(f) -> q_3, S(f) or T(f) we can infer q_3 via disjunction elimination. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall y: S(y). T(f) <-> q_3. S(f) -> q_3. Can we infer q_3 from them? If possible, name the inference rules being used at each step. ### Response: From forall y: S(y) we can infer forall y: S(y) or T(y) via universal addition. From forall y: S(y) or T(y) we can infer S(f) or T(f) via universal instantiation. From T(f) <-> q_3 we can infer T(f) -> q_3 via biconditional elimination. Finally, from S(f) -> q_3, T(f) -> q_3, S(f) or T(f) we can infer q_3 via disjunction elimination. Therefore, the answer is yes.

Consider the following premises. exists x_5: P_5(x_5) or ~R_3(x_5). forall x_5: ~P_5(x_5). forall x_5: P_5(x_5) -> R_3(x_5). Can we infer exists x_5: ~R_3(x_5) from them? If possible, name the inference rules being used at each step.

From exists x_5: P_5(x_5) or ~R_3(x_5) we can infer exists x_5: R_3(x_5) -> P_5(x_5) via existential rewriting. From exists x_5: R_3(x_5) -> P_5(x_5), forall x_5: P_5(x_5) -> R_3(x_5) we can infer exists x_5: R_3(x_5) <-> P_5(x_5) via existential biconditional introduction. Finally, from exists x_5: R_3(x_5) <-> P_5(x_5), forall x_5: ~P_5(x_5) we can infer exists x_5: ~R_3(x_5) via existential biconditional elimination. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. exists x_5: P_5(x_5) or ~R_3(x_5). forall x_5: ~P_5(x_5). forall x_5: P_5(x_5) -> R_3(x_5). Can we infer exists x_5: ~R_3(x_5) from them? If possible, name the inference rules being used at each step. ### Response: From exists x_5: P_5(x_5) or ~R_3(x_5) we can infer exists x_5: R_3(x_5) -> P_5(x_5) via existential rewriting. From exists x_5: R_3(x_5) -> P_5(x_5), forall x_5: P_5(x_5) -> R_3(x_5) we can infer exists x_5: R_3(x_5) <-> P_5(x_5) via existential biconditional introduction. Finally, from exists x_5: R_3(x_5) <-> P_5(x_5), forall x_5: ~P_5(x_5) we can infer exists x_5: ~R_3(x_5) via existential biconditional elimination. Therefore, the answer is yes.

Consider the following premises. It is sunny if and only if it is windy. It is not windy. For all x_14, x_14 will not listen to a song. Barbara is listening to a song if and only if Mary is reading a book. Can we infer the following from them? If we can, name the inference rule being used: It is not sunny and Mary is not reading a book.

From the fact that it is sunny if and only if it is windy, and that it is not windy we can infer that it is not sunny via biconditional elimination. From the fact that for all x_14, x_14 will not listen to a song we can infer that Barbara is not listening to a song via universal instantiation. From the fact that Barbara is listening to a song if and only if Mary is reading a book, and that Barbara is not listening to a song we can infer that Mary is not reading a book via biconditional elimination. Finally, from the fact that it is not sunny, and that Mary is not reading a book we can infer that it is not sunny and Mary is not reading a book via conjunction. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. It is sunny if and only if it is windy. It is not windy. For all x_14, x_14 will not listen to a song. Barbara is listening to a song if and only if Mary is reading a book. Can we infer the following from them? If we can, name the inference rule being used: It is not sunny and Mary is not reading a book. ### Response: From the fact that it is sunny if and only if it is windy, and that it is not windy we can infer that it is not sunny via biconditional elimination. From the fact that for all x_14, x_14 will not listen to a song we can infer that Barbara is not listening to a song via universal instantiation. From the fact that Barbara is listening to a song if and only if Mary is reading a book, and that Barbara is not listening to a song we can infer that Mary is not reading a book via biconditional elimination. Finally, from the fact that it is not sunny, and that Mary is not reading a book we can infer that it is not sunny and Mary is not reading a book via conjunction. Therefore, the answer is yes.

Consider the following premises. There is at least one x_12 for which x_12 will sleep or x_12 will drink water. For all x_12, x_12 is a politician. For all x_12, x_12 will drink water if and only if x_12 will cook. For all x_12, if x_12 sleeps, then x_12 is cooking. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_12 for which x_12 will cook and x_12 is a politician.

From the fact that for all x_12, x_12 will drink water if and only if x_12 will cook we can infer that for all x_12, if x_12 drinks water, then x_12 is cooking via universal biconditional elimination. From the fact that for all x_12, if x_12 sleeps, then x_12 is cooking, that for all x_12, if x_12 drinks water, then x_12 is cooking, and that there is at least one x_12 for which x_12 will sleep or x_12 will drink water we can infer that there is at least one x_12 for which x_12 will cook via existential disjunction elimination. Finally, from the fact that there is at least one x_12 for which x_12 will cook, and that for all x_12, x_12 is a politician we can infer that there is at least one x_12 for which x_12 will cook and x_12 is a politician via existential conjunction. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. There is at least one x_12 for which x_12 will sleep or x_12 will drink water. For all x_12, x_12 is a politician. For all x_12, x_12 will drink water if and only if x_12 will cook. For all x_12, if x_12 sleeps, then x_12 is cooking. Can we infer the following from them? If we can, name the inference rule being used: There is at least one x_12 for which x_12 will cook and x_12 is a politician. ### Response: From the fact that for all x_12, x_12 will drink water if and only if x_12 will cook we can infer that for all x_12, if x_12 drinks water, then x_12 is cooking via universal biconditional elimination. From the fact that for all x_12, if x_12 sleeps, then x_12 is cooking, that for all x_12, if x_12 drinks water, then x_12 is cooking, and that there is at least one x_12 for which x_12 will sleep or x_12 will drink water we can infer that there is at least one x_12 for which x_12 will cook via existential disjunction elimination. Finally, from the fact that there is at least one x_12 for which x_12 will cook, and that for all x_12, x_12 is a politician we can infer that there is at least one x_12 for which x_12 will cook and x_12 is a politician via existential conjunction. Therefore, the answer is yes.

Consider the following premises. For all x_12, x_12 will work and x_12 will play squash. There is at least one x_12 for which x_12 is bored. For all x_12, x_12 is an astronaut if and only if x_12 will work. Can we infer the following from them? If we can, name the inference rule being used: For all x_12, either x_12 is bored or x_12 is an astronaut 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_12, x_12 will work and x_12 will play squash. There is at least one x_12 for which x_12 is bored. For all x_12, x_12 is an astronaut if and only if x_12 will work. Can we infer the following from them? If we can, name the inference rule being used: For all x_12, either x_12 is bored or x_12 is an astronaut are not true. ### Response: That contradicts the premises. Therefore the answer is no.

Consider the following premises. exists z: P_1(z). forall z: ~P_1(z) or P(z). forall z: ~U(z). Can we infer forall z: ~P(z) from them? If possible, name the inference rules being used at each step.

From exists z: P_1(z) we can infer exists z: P_1(z) or U(z) via existential addition. From exists z: P_1(z) or U(z), forall z: ~P_1(z) or P(z) we can infer exists z: U(z) or P(z) via existential resolution. Finally, from exists z: U(z) or P(z), forall z: ~U(z) we can infer exists z: P(z) via existential disjunctive syllogism, which contradicts forall z: ~P(z). Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. exists z: P_1(z). forall z: ~P_1(z) or P(z). forall z: ~U(z). Can we infer forall z: ~P(z) from them? If possible, name the inference rules being used at each step. ### Response: From exists z: P_1(z) we can infer exists z: P_1(z) or U(z) via existential addition. From exists z: P_1(z) or U(z), forall z: ~P_1(z) or P(z) we can infer exists z: U(z) or P(z) via existential resolution. Finally, from exists z: U(z) or P(z), forall z: ~U(z) we can infer exists z: P(z) via existential disjunctive syllogism, which contradicts forall z: ~P(z). Therefore, the answer is no.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x, x is a sailor or x is not an astronaut. For all x, x is not a sailor. For all x, x is an astronaut or x is not a sailor.

For all x, if x were an astronaut, then x is a sailor. For all x, if x were a sailor, then x is an astronaut.

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, x is a sailor or x is not an astronaut. For all x, x is not a sailor. For all x, x is an astronaut or x is not a sailor. ### Response: For all x, if x were an astronaut, then x is a sailor. For all x, if x were a sailor, then x is an astronaut.

Consider the following premises. r_4 -> R(j_2). ~R_4(j_2). ~r_4 -> u. forall z_1: R(z_1) <-> R_4(z_1). Can we infer u from them?

From forall z_1: R(z_1) <-> R_4(z_1) we can infer R(j_2) <-> R_4(j_2). From R(j_2) <-> R_4(j_2), ~R_4(j_2) we can infer ~R(j_2). From r_4 -> R(j_2), ~R(j_2) we can infer ~r_4. Finally, from ~r_4 -> u, ~r_4 we can infer u. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. r_4 -> R(j_2). ~R_4(j_2). ~r_4 -> u. forall z_1: R(z_1) <-> R_4(z_1). Can we infer u from them? ### Response: From forall z_1: R(z_1) <-> R_4(z_1) we can infer R(j_2) <-> R_4(j_2). From R(j_2) <-> R_4(j_2), ~R_4(j_2) we can infer ~R(j_2). From r_4 -> R(j_2), ~R(j_2) we can infer ~r_4. Finally, from ~r_4 -> u, ~r_4 we can infer u. Therefore, the answer is yes.

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: ~Q_3(x_1). forall x_1: S(x_1) -> Q_3(x_1). forall x_1: S(x_1) <-> P_4(x_1).

forall x_1: ~S(x_1) can be inferred via the universal modus tollens rule. forall x_1: S(x_1) -> P_4(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: P_4(x_1) -> S(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: ~Q_3(x_1). forall x_1: S(x_1) -> Q_3(x_1). forall x_1: S(x_1) <-> P_4(x_1). ### Response: forall x_1: ~S(x_1) can be inferred via the universal modus tollens rule. forall x_1: S(x_1) -> P_4(x_1) can be inferred via the universal biconditional elimination rule. forall x_1: P_4(x_1) -> S(x_1) can be inferred via the universal biconditional elimination rule.

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 x_12: ~P_12(x_12). forall x_12: P_12(x_12) <-> U(x_12). forall x_12: T(x_12) -> U(x_12).

forall x_12: P_12(x_12) -> U(x_12) can be inferred via the universal biconditional elimination rule. forall x_12: U(x_12) -> P_12(x_12) can be inferred via the universal biconditional elimination rule. exists x_12: ~U(x_12) 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: exists x_12: ~P_12(x_12). forall x_12: P_12(x_12) <-> U(x_12). forall x_12: T(x_12) -> U(x_12). ### Response: forall x_12: P_12(x_12) -> U(x_12) can be inferred via the universal biconditional elimination rule. forall x_12: U(x_12) -> P_12(x_12) can be inferred via the universal biconditional elimination rule. exists x_12: ~U(x_12) can be inferred via the existential biconditional elimination rule.

Consider the following premises. There is at least one y_3 for which y_3 will play squash if and only if y_3 will read a book. For all y_3, y_3 will not play squash. For all y_3, y_3 will sleep if and only if y_3 will read a book. Can we infer the following from them? If we can, name the inference rule being used: There is at least one y_3 for which y_3 will not sleep.

From the fact that there is at least one y_3 for which y_3 will play squash if and only if y_3 will read a book, and that for all y_3, y_3 will not play squash we can infer that there is at least one y_3 for which y_3 will not read a book via existential biconditional elimination. From the fact that for all y_3, y_3 will sleep if and only if y_3 will read a book we can infer that for all y_3, if y_3 sleeps, then y_3 is reading a book via universal biconditional elimination. Finally, from the fact that for all y_3, if y_3 sleeps, then y_3 is reading a book, and that there is at least one y_3 for which y_3 will not read a book we can infer that there is at least one y_3 for which y_3 will not sleep via existential modus tollens. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. There is at least one y_3 for which y_3 will play squash if and only if y_3 will read a book. For all y_3, y_3 will not play squash. For all y_3, y_3 will sleep if and only if y_3 will read a book. Can we infer the following from them? If we can, name the inference rule being used: There is at least one y_3 for which y_3 will not sleep. ### Response: From the fact that there is at least one y_3 for which y_3 will play squash if and only if y_3 will read a book, and that for all y_3, y_3 will not play squash we can infer that there is at least one y_3 for which y_3 will not read a book via existential biconditional elimination. From the fact that for all y_3, y_3 will sleep if and only if y_3 will read a book we can infer that for all y_3, if y_3 sleeps, then y_3 is reading a book via universal biconditional elimination. Finally, from the fact that for all y_3, if y_3 sleeps, then y_3 is reading a book, and that there is at least one y_3 for which y_3 will not read a book we can infer that there is at least one y_3 for which y_3 will not sleep via existential modus tollens. Therefore, the answer is yes.

Consider the following premises. forall x_1: Q_1(x_1) -> U_4(x_1). forall x_1: R_3(x_1) -> Q_1(x_1). exists x_1: T(x_1) -> U_4(x_1). forall x_1: R_3(x_1) or T(x_1). Can we infer exists x_1: U_4(x_1) from them? If possible, name the inference rules being used at each step.

From forall x_1: R_3(x_1) -> Q_1(x_1), forall x_1: Q_1(x_1) -> U_4(x_1) we can infer forall x_1: R_3(x_1) -> U_4(x_1) via universal transitivity. Finally, from forall x_1: R_3(x_1) -> U_4(x_1), exists x_1: T(x_1) -> U_4(x_1), forall x_1: R_3(x_1) or T(x_1) we can infer exists x_1: U_4(x_1) via existential disjunction elimination. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_1: Q_1(x_1) -> U_4(x_1). forall x_1: R_3(x_1) -> Q_1(x_1). exists x_1: T(x_1) -> U_4(x_1). forall x_1: R_3(x_1) or T(x_1). Can we infer exists x_1: U_4(x_1) from them? If possible, name the inference rules being used at each step. ### Response: From forall x_1: R_3(x_1) -> Q_1(x_1), forall x_1: Q_1(x_1) -> U_4(x_1) we can infer forall x_1: R_3(x_1) -> U_4(x_1) via universal transitivity. Finally, from forall x_1: R_3(x_1) -> U_4(x_1), exists x_1: T(x_1) -> U_4(x_1), forall x_1: R_3(x_1) or T(x_1) we can infer exists x_1: U_4(x_1) via existential disjunction elimination. Therefore, the answer is yes.

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: t or ~q_12. w_4 or q_2. t or ~q_2. w_4 -> q_12.

q_12 -> t can be inferred via the rewriting rule. q_2 -> t 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: t or ~q_12. w_4 or q_2. t or ~q_2. w_4 -> q_12. ### Response: q_12 -> t can be inferred via the rewriting rule. q_2 -> t can be inferred via the rewriting rule.

Consider the following premises. If it is cold, then it is windy. If it is windy, then William is sad. For all x_14, x_14 is not sad. Can we infer the following from them? It is cold.

From the fact that for all x_14, x_14 is not sad we can infer that William is not sad. From the fact that if it is windy, then William is sad, and that William is not sad we can infer that it is not windy. Finally, from the fact that if it is cold, then it is windy, and that it is not windy we can infer that it is not cold, which contradicts that It is cold. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. If it is cold, then it is windy. If it is windy, then William is sad. For all x_14, x_14 is not sad. Can we infer the following from them? It is cold. ### Response: From the fact that for all x_14, x_14 is not sad we can infer that William is not sad. From the fact that if it is windy, then William is sad, and that William is not sad we can infer that it is not windy. Finally, from the fact that if it is cold, then it is windy, and that it is not windy we can infer that it is not cold, which contradicts that It is cold. Therefore, the answer is no.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? ~s or t. ~u_3. s or u. u <-> u_3.

u or t. u -> u_3. u_3 -> u. ~u.

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)? ~s or t. ~u_3. s or u. u <-> u_3. ### Response: u or t. u -> u_3. u_3 -> u. ~u.

Consider the following premises. forall x_12: Q(x_12). forall x_12: P_12(x_12) and Q_12(x_12). exists x_12: P_12(x_12) -> W(x_12). Can we infer exists x_12: W(x_12) and Q(x_12) from them? If possible, name the inference rules being used at each step.

From forall x_12: P_12(x_12) and Q_12(x_12) we can infer forall x_12: P_12(x_12) via universal simplification. From exists x_12: P_12(x_12) -> W(x_12), forall x_12: P_12(x_12) we can infer exists x_12: W(x_12) via existential modus ponens. Finally, from exists x_12: W(x_12), forall x_12: Q(x_12) we can infer exists x_12: W(x_12) and Q(x_12) via existential conjunction. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_12: Q(x_12). forall x_12: P_12(x_12) and Q_12(x_12). exists x_12: P_12(x_12) -> W(x_12). Can we infer exists x_12: W(x_12) and Q(x_12) from them? If possible, name the inference rules being used at each step. ### Response: From forall x_12: P_12(x_12) and Q_12(x_12) we can infer forall x_12: P_12(x_12) via universal simplification. From exists x_12: P_12(x_12) -> W(x_12), forall x_12: P_12(x_12) we can infer exists x_12: W(x_12) via existential modus ponens. Finally, from exists x_12: W(x_12), forall x_12: Q(x_12) we can infer exists x_12: W(x_12) and Q(x_12) via existential conjunction. Therefore, the answer is yes.

Translate the following inference to logic notation: There is at least one x for which if y_4 listens to a song, then y_4 is not climbing a mountain. For all x, y_4 will cook. For all x, y_4 will cook if and only if y_4 will climb a mountain. Therefore there is at least one x for which y_4 will not listen to a song.

exists x: P(x) -> ~Q(x). forall x: R(x). forall x: R(x) <-> Q(x). Therefore exists x: ~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 if y_4 listens to a song, then y_4 is not climbing a mountain. For all x, y_4 will cook. For all x, y_4 will cook if and only if y_4 will climb a mountain. Therefore there is at least one x for which y_4 will not listen to a song. ### Response: exists x: P(x) -> ~Q(x). forall x: R(x). forall x: R(x) <-> Q(x). Therefore exists x: ~P(x).

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: ~t. t. u -> q.

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)? Name the inference rule being used: ~t. t. u -> q. ### Response: Since the premises are contradictory, we can infer anything from them.

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_8 for which if x_8 listens to a song, then x_8 is a carpenter. For all x_8, if x_8 were a taxi driver, then x_8 is listening to a song. For all x_8, x_8 will listen to a song. For all x_8, if x_8 were a carpenter, then x_8 is a taxi driver.

There is at least one x_8 for which x_8 is a carpenter. For all x_8, if x_8 were a carpenter, then x_8 is listening to a song. There is at least one x_8 for which if x_8 listens to a song, then x_8 is a taxi driver. There is at least one x_8 for which if x_8 were a taxi driver, then x_8 is a carpenter.

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_8 for which if x_8 listens to a song, then x_8 is a carpenter. For all x_8, if x_8 were a taxi driver, then x_8 is listening to a song. For all x_8, x_8 will listen to a song. For all x_8, if x_8 were a carpenter, then x_8 is a taxi driver. ### Response: There is at least one x_8 for which x_8 is a carpenter. For all x_8, if x_8 were a carpenter, then x_8 is listening to a song. There is at least one x_8 for which if x_8 listens to a song, then x_8 is a taxi driver. There is at least one x_8 for which if x_8 were a taxi driver, then x_8 is a carpenter.

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_13. q. s <-> q_13.

q_13 and q can be inferred via the conjunction rule. q and q_13 can be inferred via the conjunction rule. s -> q_13 can be inferred via the biconditional elimination rule. q_13 -> s 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_13. q. s <-> q_13. ### Response: q_13 and q can be inferred via the conjunction rule. q and q_13 can be inferred via the conjunction rule. s -> q_13 can be inferred via the biconditional elimination rule. q_13 -> s can be inferred via the biconditional elimination rule.

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: It is snowing. Jessica is cooking if and only if it is snowing.

If Jessica cooks, then it is snowing can be inferred via the biconditional elimination rule. If it snows, then Jessica is cooking 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: It is snowing. Jessica is cooking if and only if it is snowing. ### Response: If Jessica cooks, then it is snowing can be inferred via the biconditional elimination rule. If it snows, then Jessica is cooking can be inferred via the biconditional elimination rule.

Consider the following premises. Robert is not running. Robert is running if and only if it is overcast. If it is windy, then it is overcast. If John were a cashier, then it is windy. Can we infer the following from them? If we can, name the inference rule being used: It is late.

We cannot infer that from the premises. Therefore the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. Robert is not running. Robert is running if and only if it is overcast. If it is windy, then it is overcast. If John were a cashier, then it is windy. Can we infer the following from them? If we can, name the inference rule being used: It is late. ### Response: We cannot infer that from the premises. Therefore the answer is no.

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: W(z). forall z: W(z) -> S(z). exists z: U_3(z).

forall z: S(z) can be inferred via the universal modus ponens rule. exists z: U_3(z) and W(z) can be inferred via the existential conjunction rule. exists z: W(z) and U_3(z) can be inferred via the existential conjunction 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: W(z). forall z: W(z) -> S(z). exists z: U_3(z). ### Response: forall z: S(z) can be inferred via the universal modus ponens rule. exists z: U_3(z) and W(z) can be inferred via the existential conjunction rule. exists z: W(z) and U_3(z) can be inferred via the existential conjunction rule.

Consider the following premises. p_2 or ~p_10. p_10 and r_1. Can we infer s_4 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. p_2 or ~p_10. p_10 and r_1. Can we infer s_4 from them? ### Response: We cannot infer that from the premises. Therefore the answer is no.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all x_15, x_15 is a server. For all x_15, x_15 will listen to a song and x_15 will go running. It is raining.

For all x_15, x_15 will listen to a song. For all x_15, x_15 will go 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_15, x_15 is a server. For all x_15, x_15 will listen to a song and x_15 will go running. It is raining. ### Response: For all x_15, x_15 will listen to a song. For all x_15, x_15 will go running.

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) -> P_9(x_9). forall x_9: P_9(x_9) -> S(x_9). forall x_9: P_9(x_9).

forall x_9: S(x_9). forall x_9: S(x_9) <-> P_9(x_9). forall x_9: P_9(x_9) <-> S(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) -> P_9(x_9). forall x_9: P_9(x_9) -> S(x_9). forall x_9: P_9(x_9). ### Response: forall x_9: S(x_9). forall x_9: S(x_9) <-> P_9(x_9). forall x_9: P_9(x_9) <-> S(x_9).

Consider the following premises. Jessica is not an electrician. For all x_14, x_14 will listen to a song or x_14 is an electrician. For all x_14, x_14 will not listen to a song. Can we infer the following from them? If we can, name the inference rule being used: It is not sunny.

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. Jessica is not an electrician. For all x_14, x_14 will listen to a song or x_14 is an electrician. For all x_14, x_14 will not listen to a song. Can we infer the following from them? If we can, name the inference rule being used: It is not sunny. ### Response: The premises are contradictory and we can infer anything from them. Therefore, the answer is yes.

Consider the following premises. ~q_15. p_13 <-> q. q -> q_15. t_2 or p_13. Can we infer u 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_15. p_13 <-> q. q -> q_15. t_2 or p_13. Can we infer u 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.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? Elisabeth is a mechanic. Joseph is poor.

Elisabeth is a mechanic and Joseph is poor. Joseph is poor and Elisabeth is a mechanic.

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 a mechanic. Joseph is poor. ### Response: Elisabeth is a mechanic and Joseph is poor. Joseph is poor and Elisabeth is a mechanic.

Translate the following inference to logic notation: For all x, x will work. For all x, x will go running. For all x, if x works, then x is a scientist. Therefore for all x, x will go running.

forall x: P(x). forall x: Q(x). forall x: P(x) -> R(x). Therefore forall x: Q(x).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, x will work. For all x, x will go running. For all x, if x works, then x is a scientist. Therefore for all x, x will go running. ### Response: forall x: P(x). forall x: Q(x). forall x: P(x) -> R(x). Therefore forall x: Q(x).

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall z: Q_2(z) or ~R(z). forall z: ~Q_2(z). exists z: Q_2(z) -> R(z).

forall z: R(z) -> Q_2(z).

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: Q_2(z) or ~R(z). forall z: ~Q_2(z). exists z: Q_2(z) -> R(z). ### Response: forall z: R(z) -> Q_2(z).

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 -> q_4. q or ~q_4.

q_4 -> q 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 -> q_4. q or ~q_4. ### Response: q_4 -> q can be inferred via the rewriting rule.

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_13, x_13 will take a plane. For all x_13, x_13 will climb a mountain. For all x_13, if x_13 climbs a mountain, then x_13 is a sailor.

For all x_13, x_13 is a sailor can be inferred via the universal modus ponens rule. For all x_13, x_13 will take a plane and x_13 will climb a mountain can be inferred via the universal conjunction rule. For all x_13, x_13 will climb a mountain and x_13 will take a plane can be inferred via the universal conjunction 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_13, x_13 will take a plane. For all x_13, x_13 will climb a mountain. For all x_13, if x_13 climbs a mountain, then x_13 is a sailor. ### Response: For all x_13, x_13 is a sailor can be inferred via the universal modus ponens rule. For all x_13, x_13 will take a plane and x_13 will climb a mountain can be inferred via the universal conjunction rule. For all x_13, x_13 will climb a mountain and x_13 will take a plane can be inferred via the universal conjunction rule.

Translate the following inference to logic notation: There is at least one x for which x_11 will listen to a song or x_11 is a writer. For all x, x_11 is not a writer. For all x, x_11 will not listen to a song or x_11 is a soldier. For all x, if x_11 were a soldier, then x_11 is bored. Therefore there is at least one x for which x_11 is bored.

exists x: P(x) or Q(x). forall x: ~Q(x). forall x: ~P(x) or R(x). forall x: R(x) -> S(x). Therefore exists x: S(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_11 will listen to a song or x_11 is a writer. For all x, x_11 is not a writer. For all x, x_11 will not listen to a song or x_11 is a soldier. For all x, if x_11 were a soldier, then x_11 is bored. Therefore there is at least one x for which x_11 is bored. ### Response: exists x: P(x) or Q(x). forall x: ~Q(x). forall x: ~P(x) or R(x). forall x: R(x) -> S(x). Therefore exists x: S(x).

Consider the following premises. exists x_12: Q_12(x_12) -> T_4(x_12). forall x_12: P_12(x_12) -> T_4(x_12). forall x_12: P_2(x_12) or Q_2(x_12). forall x_12: P_12(x_12) or Q_12(x_12). forall x_12: ~P_2(x_12). Can we infer forall x_12: ~(T_4(x_12) and Q_2(x_12)) from them? If possible, name the inference rules being used at each step.

From forall x_12: P_12(x_12) -> T_4(x_12), exists x_12: Q_12(x_12) -> T_4(x_12), forall x_12: P_12(x_12) or Q_12(x_12) we can infer exists x_12: T_4(x_12) via existential disjunction elimination. From forall x_12: P_2(x_12) or Q_2(x_12), forall x_12: ~P_2(x_12) we can infer forall x_12: Q_2(x_12) via universal disjunctive syllogism. Finally, from exists x_12: T_4(x_12), forall x_12: Q_2(x_12) we can infer exists x_12: T_4(x_12) and Q_2(x_12) via existential conjunction, which contradicts forall x_12: ~(T_4(x_12) and Q_2(x_12)). Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. exists x_12: Q_12(x_12) -> T_4(x_12). forall x_12: P_12(x_12) -> T_4(x_12). forall x_12: P_2(x_12) or Q_2(x_12). forall x_12: P_12(x_12) or Q_12(x_12). forall x_12: ~P_2(x_12). Can we infer forall x_12: ~(T_4(x_12) and Q_2(x_12)) from them? If possible, name the inference rules being used at each step. ### Response: From forall x_12: P_12(x_12) -> T_4(x_12), exists x_12: Q_12(x_12) -> T_4(x_12), forall x_12: P_12(x_12) or Q_12(x_12) we can infer exists x_12: T_4(x_12) via existential disjunction elimination. From forall x_12: P_2(x_12) or Q_2(x_12), forall x_12: ~P_2(x_12) we can infer forall x_12: Q_2(x_12) via universal disjunctive syllogism. Finally, from exists x_12: T_4(x_12), forall x_12: Q_2(x_12) we can infer exists x_12: T_4(x_12) and Q_2(x_12) via existential conjunction, which contradicts forall x_12: ~(T_4(x_12) and Q_2(x_12)). Therefore, the answer is no.

Consider the following premises. t_3. t_4 or q. q_19. ~t_4. Can we infer ~q from them? If possible, name the inference rules being used at each step.

From t_3, q_19 we can infer t_3 and q_19 via conjunction. Finally, from t_4 or q, ~t_4 we can infer q via disjunctive syllogism, which contradicts ~q. Therefore, the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. t_3. t_4 or q. q_19. ~t_4. Can we infer ~q from them? If possible, name the inference rules being used at each step. ### Response: From t_3, q_19 we can infer t_3 and q_19 via conjunction. Finally, from t_4 or q, ~t_4 we can infer q via disjunctive syllogism, which contradicts ~q. Therefore, the answer is no.

Consider the following premises. forall y_1: T_4(y_1) or ~Q(y_1). forall y_1: T(y_1) -> S_3(y_1). exists y_1: T(y_1) <-> W(y_1). forall y_1: T_4(y_1) -> W(y_1). Can we infer exists y_1: Q(y_1) -> S_3(y_1) from them? If possible, name the inference rules being used at each step.

From exists y_1: T(y_1) <-> W(y_1) we can infer exists y_1: W(y_1) -> T(y_1) via existential biconditional elimination. From exists y_1: W(y_1) -> T(y_1), forall y_1: T(y_1) -> S_3(y_1) we can infer exists y_1: W(y_1) -> S_3(y_1) via existential transitivity. From forall y_1: T_4(y_1) -> W(y_1), exists y_1: W(y_1) -> S_3(y_1) we can infer exists y_1: T_4(y_1) -> S_3(y_1) via existential transitivity. From forall y_1: T_4(y_1) or ~Q(y_1) we can infer forall y_1: Q(y_1) -> T_4(y_1) via universal rewriting. Finally, from forall y_1: Q(y_1) -> T_4(y_1), exists y_1: T_4(y_1) -> S_3(y_1) we can infer exists y_1: Q(y_1) -> S_3(y_1) via existential transitivity. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall y_1: T_4(y_1) or ~Q(y_1). forall y_1: T(y_1) -> S_3(y_1). exists y_1: T(y_1) <-> W(y_1). forall y_1: T_4(y_1) -> W(y_1). Can we infer exists y_1: Q(y_1) -> S_3(y_1) from them? If possible, name the inference rules being used at each step. ### Response: From exists y_1: T(y_1) <-> W(y_1) we can infer exists y_1: W(y_1) -> T(y_1) via existential biconditional elimination. From exists y_1: W(y_1) -> T(y_1), forall y_1: T(y_1) -> S_3(y_1) we can infer exists y_1: W(y_1) -> S_3(y_1) via existential transitivity. From forall y_1: T_4(y_1) -> W(y_1), exists y_1: W(y_1) -> S_3(y_1) we can infer exists y_1: T_4(y_1) -> S_3(y_1) via existential transitivity. From forall y_1: T_4(y_1) or ~Q(y_1) we can infer forall y_1: Q(y_1) -> T_4(y_1) via universal rewriting. Finally, from forall y_1: Q(y_1) -> T_4(y_1), exists y_1: T_4(y_1) -> S_3(y_1) we can infer exists y_1: Q(y_1) -> S_3(y_1) via existential transitivity. Therefore, the answer is yes.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all z_4, if z_4 plays a game, then z_4 is writing a letter. There is at least one z_4 for which if z_4 writes a letter, then z_4 is playing a game. For all z_4, if z_4 plays a game, then z_4 is working. For all z_4, z_4 will not work.

For all z_4, z_4 will not play a game. There is at least one z_4 for which if z_4 writes a letter, then z_4 is working. There is at least one z_4 for which z_4 will write a letter if and only if z_4 will play a game. There is at least one z_4 for which z_4 will play a game if and only if z_4 will write a letter.

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 z_4, if z_4 plays a game, then z_4 is writing a letter. There is at least one z_4 for which if z_4 writes a letter, then z_4 is playing a game. For all z_4, if z_4 plays a game, then z_4 is working. For all z_4, z_4 will not work. ### Response: For all z_4, z_4 will not play a game. There is at least one z_4 for which if z_4 writes a letter, then z_4 is working. There is at least one z_4 for which z_4 will write a letter if and only if z_4 will play a game. There is at least one z_4 for which z_4 will play a game if and only if z_4 will write a letter.

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_4 for which if x_4 goes running, then x_4 is a carpenter. For all x_4, if x_4 were a soldier, then x_4 is running. For all x_4, x_4 will cook or x_4 is a soldier. For all x_4, if x_4 cooks, then x_4 is running.

There is at least one x_4 for which if x_4 were a soldier, then x_4 is a carpenter can be inferred via the existential transitivity rule. There is at least one x_4 for which if x_4 cooks, then x_4 is a carpenter can be inferred via the existential transitivity rule. For all x_4, x_4 will go running can be inferred via the universal disjunction 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 x_4 for which if x_4 goes running, then x_4 is a carpenter. For all x_4, if x_4 were a soldier, then x_4 is running. For all x_4, x_4 will cook or x_4 is a soldier. For all x_4, if x_4 cooks, then x_4 is running. ### Response: There is at least one x_4 for which if x_4 were a soldier, then x_4 is a carpenter can be inferred via the existential transitivity rule. There is at least one x_4 for which if x_4 cooks, then x_4 is a carpenter can be inferred via the existential transitivity rule. For all x_4, x_4 will go running can be inferred via the universal disjunction elimination rule.

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_7, x_7 will not work. For all x_7, if x_7 plays squash, then x_7 is a janitor. For all x_7, x_7 will play squash if and only if x_7 is bored. There is at least one x_7 for which if x_7 were a janitor, then x_7 is playing squash. For all x_7, if x_7 were bored, then x_7 is working.

For all x_7, x_7 is not bored can be inferred via the universal modus tollens rule. There is at least one x_7 for which x_7 is a janitor if and only if x_7 will play squash can be inferred via the existential biconditional introduction rule. There is at least one x_7 for which x_7 will play squash if and only if x_7 is a janitor can be inferred via the existential biconditional introduction rule. For all x_7, if x_7 plays squash, then x_7 is bored can be inferred via the universal biconditional elimination rule. For all x_7, if x_7 were bored, then x_7 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_7, x_7 will not work. For all x_7, if x_7 plays squash, then x_7 is a janitor. For all x_7, x_7 will play squash if and only if x_7 is bored. There is at least one x_7 for which if x_7 were a janitor, then x_7 is playing squash. For all x_7, if x_7 were bored, then x_7 is working. ### Response: For all x_7, x_7 is not bored can be inferred via the universal modus tollens rule. There is at least one x_7 for which x_7 is a janitor if and only if x_7 will play squash can be inferred via the existential biconditional introduction rule. There is at least one x_7 for which x_7 will play squash if and only if x_7 is a janitor can be inferred via the existential biconditional introduction rule. For all x_7, if x_7 plays squash, then x_7 is bored can be inferred via the universal biconditional elimination rule. For all x_7, if x_7 were bored, then x_7 is playing squash can be inferred via the universal biconditional elimination rule.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_7: U(x_7). exists x_7: U_1(x_7). forall x_7: U(x_7) <-> U_2(x_7).

exists x_7: U_1(x_7) and U(x_7). exists x_7: U(x_7) and U_1(x_7). forall x_7: U(x_7) -> U_2(x_7). forall x_7: U_2(x_7) -> U(x_7).

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_7: U(x_7). exists x_7: U_1(x_7). forall x_7: U(x_7) <-> U_2(x_7). ### Response: exists x_7: U_1(x_7) and U(x_7). exists x_7: U(x_7) and U_1(x_7). forall x_7: U(x_7) -> U_2(x_7). forall x_7: U_2(x_7) -> U(x_7).

Consider the following premises. forall x_16: W_4(x_16) and S(x_16). forall x_16: W_4(x_16) -> U(x_16). u_2. Can we infer ~u_2 from them?

That contradicts the premises. Therefore the answer is no.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_16: W_4(x_16) and S(x_16). forall x_16: W_4(x_16) -> U(x_16). u_2. Can we infer ~u_2 from them? ### Response: That contradicts the premises. Therefore the answer is no.

Translate the following inference to logic notation: For all x, x_14 is a police officer or x_14 will play tennis. For all x, x_14 is not a police officer. For all x, x_14 will write a letter or x_14 will not play tennis. For all x, x_14 will write a letter if and only if x_14 will take a plane. Therefore for all x, x_14 will write a letter.

forall x: P(x) or Q(x). forall x: ~P(x). forall x: R(x) or ~Q(x). forall x: R(x) <-> S(x). Therefore forall x: R(x).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, x_14 is a police officer or x_14 will play tennis. For all x, x_14 is not a police officer. For all x, x_14 will write a letter or x_14 will not play tennis. For all x, x_14 will write a letter if and only if x_14 will take a plane. Therefore for all x, x_14 will write a letter. ### Response: forall x: P(x) or Q(x). forall x: ~P(x). forall x: R(x) or ~Q(x). forall x: R(x) <-> S(x). Therefore forall x: R(x).

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). p_12 -> ~q_12. q_12.

~p_12 can be inferred via the 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: P(x). p_12 -> ~q_12. q_12. ### Response: ~p_12 can be inferred via the modus tollens rule.

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? forall x_8: P_8(x_8). forall x_8: P_8(x_8) -> Q_8(x_8). forall x_8: Q_8(x_8) -> R_8(x_8). forall x_8: Q_3(x_8).

forall x_8: Q_8(x_8). forall x_8: P_8(x_8) -> R_8(x_8). forall x_8: P_8(x_8) and Q_3(x_8). forall x_8: Q_3(x_8) and P_8(x_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)? forall x_8: P_8(x_8). forall x_8: P_8(x_8) -> Q_8(x_8). forall x_8: Q_8(x_8) -> R_8(x_8). forall x_8: Q_3(x_8). ### Response: forall x_8: Q_8(x_8). forall x_8: P_8(x_8) -> R_8(x_8). forall x_8: P_8(x_8) and Q_3(x_8). forall x_8: Q_3(x_8) and P_8(x_8).

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? exists x_2: Q_2(x_2) -> R(x_2). forall x_2: R_2(x_2) -> R(x_2). forall x_2: P_2(x_2) or Q_2(x_2). forall x_2: ~P_2(x_2) or R_2(x_2).

forall x_2: Q_2(x_2) or R_2(x_2).

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)? exists x_2: Q_2(x_2) -> R(x_2). forall x_2: R_2(x_2) -> R(x_2). forall x_2: P_2(x_2) or Q_2(x_2). forall x_2: ~P_2(x_2) or R_2(x_2). ### Response: forall x_2: Q_2(x_2) or R_2(x_2).

Consider the following premises. r_4 -> u_2. p or w_3. ~p. w_3 -> u_2. Can we infer p_1 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. r_4 -> u_2. p or w_3. ~p. w_3 -> u_2. Can we infer p_1 from them? ### Response: We cannot infer that from the premises. Therefore the answer is no.

Consider the following premises. forall x_9: P_6(x_9) <-> T(x_9). forall x_9: P_4(x_9) -> Q_9(x_9). exists x_9: ~P_6(x_9). forall x_9: Q_9(x_9) -> T(x_9). Can we infer exists x_9: ~P_4(x_9) from them? If possible, name the inference rules being used at each step.

From forall x_9: P_4(x_9) -> Q_9(x_9), forall x_9: Q_9(x_9) -> T(x_9) we can infer forall x_9: P_4(x_9) -> T(x_9) via universal transitivity. From forall x_9: P_6(x_9) <-> T(x_9), exists x_9: ~P_6(x_9) we can infer exists x_9: ~T(x_9) via existential biconditional elimination. Finally, from forall x_9: P_4(x_9) -> T(x_9), exists x_9: ~T(x_9) we can infer exists x_9: ~P_4(x_9) via existential modus tollens. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. forall x_9: P_6(x_9) <-> T(x_9). forall x_9: P_4(x_9) -> Q_9(x_9). exists x_9: ~P_6(x_9). forall x_9: Q_9(x_9) -> T(x_9). Can we infer exists x_9: ~P_4(x_9) from them? If possible, name the inference rules being used at each step. ### Response: From forall x_9: P_4(x_9) -> Q_9(x_9), forall x_9: Q_9(x_9) -> T(x_9) we can infer forall x_9: P_4(x_9) -> T(x_9) via universal transitivity. From forall x_9: P_6(x_9) <-> T(x_9), exists x_9: ~P_6(x_9) we can infer exists x_9: ~T(x_9) via existential biconditional elimination. Finally, from forall x_9: P_4(x_9) -> T(x_9), exists x_9: ~T(x_9) we can infer exists x_9: ~P_4(x_9) via existential modus tollens. Therefore, the answer is yes.

Consider the following premises. forall x_6: P_6(x_6). ~W(g_3) or w. forall x_6: ~P_6(x_6) or R_5(x_6). Can we infer R_5(g_3) 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. forall x_6: P_6(x_6). ~W(g_3) or w. forall x_6: ~P_6(x_6) or R_5(x_6). Can we infer R_5(g_3) from them? ### Response: We cannot infer that from the premises. Therefore the answer is no.

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_12: Q_5(x_12) -> S_1(x_12). forall x_12: Q_12(x_12) -> R_12(x_12). forall x_12: P_12(x_12) or Q_12(x_12). forall x_12: P_12(x_12) -> R_12(x_12). exists x_12: T(x_12) -> ~S_1(x_12). forall x_12: R_12(x_12) -> Q_5(x_12).

forall x_12: Q_12(x_12) -> Q_5(x_12) can be inferred via the universal transitivity rule. forall x_12: P_12(x_12) -> Q_5(x_12) can be inferred via the universal transitivity rule. forall x_12: R_12(x_12) -> S_1(x_12) can be inferred via the universal transitivity rule. forall x_12: R_12(x_12) can be inferred via the universal disjunction 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_12: Q_5(x_12) -> S_1(x_12). forall x_12: Q_12(x_12) -> R_12(x_12). forall x_12: P_12(x_12) or Q_12(x_12). forall x_12: P_12(x_12) -> R_12(x_12). exists x_12: T(x_12) -> ~S_1(x_12). forall x_12: R_12(x_12) -> Q_5(x_12). ### Response: forall x_12: Q_12(x_12) -> Q_5(x_12) can be inferred via the universal transitivity rule. forall x_12: P_12(x_12) -> Q_5(x_12) can be inferred via the universal transitivity rule. forall x_12: R_12(x_12) -> S_1(x_12) can be inferred via the universal transitivity rule. forall x_12: R_12(x_12) can be inferred via the universal disjunction elimination rule.

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, z is a janitor if and only if z will play squash. If Elisabeth were a janitor, then Linda is rich. Linda is not rich.

Elisabeth is not a janitor can be inferred via the modus tollens rule. For all z, if z were a janitor, then z is playing squash can be inferred via the universal biconditional elimination rule. For all z, if z plays squash, then z is a janitor 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, z is a janitor if and only if z will play squash. If Elisabeth were a janitor, then Linda is rich. Linda is not rich. ### Response: Elisabeth is not a janitor can be inferred via the modus tollens rule. For all z, if z were a janitor, then z is playing squash can be inferred via the universal biconditional elimination rule. For all z, if z plays squash, then z is a janitor can be inferred via the universal biconditional elimination rule.

Translate the following inference to logic notation: If it snows, then Jessica is sleeping. Thomas is taking a plane if and only if it is snowing. It is snowing or Jessica is not sleeping. Jessica is not sleeping. Therefore Thomas is not taking a plane.

p -> q. r <-> p. p or ~q. ~q. Therefore ~r.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: If it snows, then Jessica is sleeping. Thomas is taking a plane if and only if it is snowing. It is snowing or Jessica is not sleeping. Jessica is not sleeping. Therefore Thomas is not taking a plane. ### Response: p -> q. r <-> p. p or ~q. ~q. Therefore ~r.

Consider the following premises. For all x_5, if x_5 were a police officer, then x_5 is not an astronaut. For all x_5, if x_5 drinks water, then x_5 is an astronaut. For all x_5, x_5 will drink water. For all x_5, x_5 will go running if and only if x_5 is a police officer. Can we infer the following from them? If we can, name the inference rule being used: For all x_5, if x_5 drinks water, then x_5 is an astronaut.

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, if x_5 were a police officer, then x_5 is not an astronaut. For all x_5, if x_5 drinks water, then x_5 is an astronaut. For all x_5, x_5 will drink water. For all x_5, x_5 will go running if and only if x_5 is a police officer. Can we infer the following from them? If we can, name the inference rule being used: For all x_5, if x_5 drinks water, then x_5 is an astronaut. ### Response: That is one of the premises. Therefore, the answer is yes.

Consider the following premises. If David were sad, then Thomas is drinking water. For all x_14, if x_14 drinks water, then x_14 is sleeping. David is sad. Can we infer the following from them? David is sad.

That is one of the premises. Therefore, the answer is yes.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Consider the following premises. If David were sad, then Thomas is drinking water. For all x_14, if x_14 drinks water, then x_14 is sleeping. David is sad. Can we infer the following from them? David is sad. ### Response: That is one of the premises. Therefore, the answer is yes.

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_11, x_11 is a politician if and only if x_11 is bored. There is at least one x_11 for which if x_11 were a bartender, then x_11 is a politician. For all x_11, if x_11 were an office clerk, then x_11 is a lawyer. For all x_11, if x_11 were bored, then x_11 is an office clerk.

For all x_11, if x_11 were bored, then x_11 is a lawyer can be inferred via the universal transitivity rule. For all x_11, if x_11 were a politician, then x_11 is bored can be inferred via the universal biconditional elimination rule. For all x_11, if x_11 were bored, then x_11 is a politician 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_11, x_11 is a politician if and only if x_11 is bored. There is at least one x_11 for which if x_11 were a bartender, then x_11 is a politician. For all x_11, if x_11 were an office clerk, then x_11 is a lawyer. For all x_11, if x_11 were bored, then x_11 is an office clerk. ### Response: For all x_11, if x_11 were bored, then x_11 is a lawyer can be inferred via the universal transitivity rule. For all x_11, if x_11 were a politician, then x_11 is bored can be inferred via the universal biconditional elimination rule. For all x_11, if x_11 were bored, then x_11 is a politician can be inferred via the universal biconditional elimination rule.

Translate the following inference to logic notation: For all x, x is a politician or x is a taxi driver. For all x, x is fast. For all x, x is not a politician. Therefore for all x, x is a taxi driver and x is fast.

forall x: P(x) or Q(x). forall x: R(x). forall x: ~P(x). Therefore forall x: Q(x) and R(x).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, x is a politician or x is a taxi driver. For all x, x is fast. For all x, x is not a politician. Therefore for all x, x is a taxi driver and x is fast. ### Response: forall x: P(x) or Q(x). forall x: R(x). forall x: ~P(x). Therefore forall x: Q(x) and R(x).

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 is fast. For all x_5, x_5 will work and x_5 will read a book. For all x_5, if x_5 were fast, then x_5 is bored.

For all x_5, x_5 is bored can be inferred via the universal modus ponens rule. For all x_5, x_5 will work can be inferred via the universal simplification rule. For all x_5, x_5 will read a book can be inferred via the universal simplification 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 is fast. For all x_5, x_5 will work and x_5 will read a book. For all x_5, if x_5 were fast, then x_5 is bored. ### Response: For all x_5, x_5 is bored can be inferred via the universal modus ponens rule. For all x_5, x_5 will work can be inferred via the universal simplification rule. For all x_5, x_5 will read a book can be inferred via the universal simplification rule.

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_9: Q_9(x_9) -> P_1(x_9). forall x_9: T_3(x_9) -> Q_9(x_9). forall x_9: P_1(x_9) <-> Q_7(x_9). forall x_9: ~Q_7(x_9).

forall x_9: T_3(x_9) -> P_1(x_9) can be inferred via the universal transitivity rule. forall x_9: P_1(x_9) -> Q_7(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_7(x_9) -> P_1(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~P_1(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: forall x_9: Q_9(x_9) -> P_1(x_9). forall x_9: T_3(x_9) -> Q_9(x_9). forall x_9: P_1(x_9) <-> Q_7(x_9). forall x_9: ~Q_7(x_9). ### Response: forall x_9: T_3(x_9) -> P_1(x_9) can be inferred via the universal transitivity rule. forall x_9: P_1(x_9) -> Q_7(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: Q_7(x_9) -> P_1(x_9) can be inferred via the universal biconditional elimination rule. forall x_9: ~P_1(x_9) can be inferred via the universal biconditional elimination rule.

Translate the following inference to logic notation: It is sunny. If it is sunny, then Mary is driving a car. Robert is a bartender if and only if Mary is driving a car. Therefore Mary is driving a car.

p. p -> q. r <-> q. Therefore q.

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: It is sunny. If it is sunny, then Mary is driving a car. Robert is a bartender if and only if Mary is driving a car. Therefore Mary is driving a car. ### Response: p. p -> q. r <-> q. Therefore q.

Translate the following inference to logic notation: For all x, if x_9 were a poet, then x_9 is writing a letter. For all x, x_9 is a poet or x_9 will not write a letter. There is at least one x for which if x_9 were a poet, then x_9 is writing a letter. Therefore there is at least one x for which if x_9 writes a letter, then x_9 is a poet.

forall x: P(x) -> Q(x). forall x: P(x) or ~Q(x). exists x: P(x) -> Q(x). Therefore exists x: Q(x) -> P(x).

LogicInference Dataset e

### Instruction: Answer the following. ### Input: Translate the following inference to logic notation: For all x, if x_9 were a poet, then x_9 is writing a letter. For all x, x_9 is a poet or x_9 will not write a letter. There is at least one x for which if x_9 were a poet, then x_9 is writing a letter. Therefore there is at least one x for which if x_9 writes a letter, then x_9 is a poet. ### Response: forall x: P(x) -> Q(x). forall x: P(x) or ~Q(x). exists x: P(x) -> Q(x). Therefore exists x: Q(x) -> P(x).

What can be inferred from the following premises in a single inference step (ignoring inferences that add new predicates or constants)? For all z, if z were a bartender, then z is happy. For all z, if z were happy, then z is a bartender. For all z, z is happy if and only if z is poor. For all z, z is not poor.

For all z, z is a bartender if and only if z is happy. For all z, z is happy if and only if z is a bartender. For all z, if z were happy, then z is poor. For all z, if z were poor, then z is happy. For all z, z is not happy.

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 z, if z were a bartender, then z is happy. For all z, if z were happy, then z is a bartender. For all z, z is happy if and only if z is poor. For all z, z is not poor. ### Response: For all z, z is a bartender if and only if z is happy. For all z, z is happy if and only if z is a bartender. For all z, if z were happy, then z is poor. For all z, if z were poor, then z is happy. For all z, z is not happy.