Propositional Logic Practice Questions

A. $a \vee b \to b \wedge c$

B. $a \wedge b \to b \vee c$

C. $a \vee b \to \left(b \to c \right)$

D. $a \to b \to \left(b \to c \right)$

A. $(\forall (x)) P(x) \vee (\forall(x))Q(x) \implies (\forall (x)) (P(x) \vee Q(x))$

B. $(\exists (x)) P(x) \wedge (\exists (x))Q(x) \implies (\exists (x)) (P(x) \wedge Q(x))$

C. $(\forall (x)) (P(x) \vee Q(x)) \implies (\forall (x)) P(x) \vee (\forall (x)) Q(x)$

D. $(\exists (x)) (P(x) \vee Q(x)) \implies \sim (\forall (x)) P(x) \vee (\exists (x)) Q(x)$

A. Both $F_1$ and $F_2$ are tautologies

B. The conjunction $F_1 \land F_2$ is not satisfiable

C. Neither is tautologous

D. Neither is satisfiable

E. None of the above.

A. $\left(P\Rightarrow Q\right) {\wedge} \left(Q \Rightarrow R\right) \Rightarrow \left(P \Rightarrow R\right)$

B. $\left(P\Rightarrow Q\right) \Rightarrow \left( \neg P \Rightarrow \neg Q\right)$

C. $\left(P{\wedge} \left(\neg P \vee \neg Q\right)\right) \Rightarrow Q$

D. $\left(P \Rightarrow R\right) \vee \left(Q \Rightarrow R\right) \Rightarrow \left(\left(P \vee Q \right) \Rightarrow R\right)$

A. $P \rightarrow Q$

B. $\neg Q \rightarrow \neg P$

C. $\neg P \vee Q$

D. $\neg Q \rightarrow P$