$P$ and $Q$ are distinct atomic sentences.
Now let’s take $P = T, Q = F$; $P = F, Q = T$; $P = T, Q = T$; $P = F, Q = F$ [4 Cases]
$\therefore \neg P \vee Q \equiv F \vee F \equiv F$;
$\neg P \vee Q \equiv T \vee T \equiv T$;
$\neg P \vee Q \equiv F \vee T \equiv T$;
$\neg P \vee Q \equiv T \vee F \equiv T$
Option by option:
- $\neg Q → \neg P$
The sentence $\neg P \vee Q$ $tautologically$ $implies$ sentence $\neg Q → \neg P$
Case 1: $(\neg P \vee Q) → (\neg Q → \neg P) \equiv F → F \equiv T$
Case 2: $(\neg P \vee Q) → (\neg Q → \neg P) \equiv T → T \equiv T$
Case 3: $(\neg P \vee Q) → (\neg Q → \neg P) \equiv T → T \equiv T$
Case 4: $(\neg P \vee Q) → (\neg Q → \neg P) \equiv T → T \equiv T$
- $Q → P$
The sentence $\neg P \vee Q$ $NOT$ $tautologically$ $implies$ sentence $Q → P$
Case 1: $(\neg P \vee Q) → (Q → P) \equiv F → T \equiv T$
Case 2: $(\neg P \vee Q) → (Q → P) \equiv T → F \equiv F$
Case 3: $(\neg P \vee Q) → (Q → P) \equiv T → T \equiv T$
Case 4: $(\neg P \vee Q) → (Q → P) \equiv T → T \equiv T$
- $P → Q$
The sentence $\neg P \vee Q$ $tautologically$ $implies$ sentence $P → Q$
Case 1: $(\neg P \vee Q) → (P → Q) \equiv F → F \equiv T$
Case 2: $(\neg P \vee Q) → (P → Q) \equiv T → T \equiv T$
Case 3: $(\neg P \vee Q) → (P → Q) \equiv T → T \equiv T$
Case 4: $(\neg P \vee Q) → (P → Q) \equiv T → T \equiv T$
- $\neg P \wedge Q$
The sentence $\neg P \vee Q$ $NOT$ $tautologically$ $implies$ sentence $\neg P \wedge Q$
Case 1: $(\neg P \vee Q) → (\neg P \wedge Q) \equiv F → F \equiv T$
Case 2: $(\neg P \vee Q) → (\neg P \wedge Q) \equiv T → T \equiv T$
Case 3: $(\neg P \vee Q) → (\neg P \wedge Q) \equiv T → F \equiv F$
Case 4: $(\neg P \vee Q) → (\neg P \wedge Q) \equiv T → F \equiv F$
$Ans: A;C$
(Can be solved without making cases as $\neg P \vee Q \equiv P \rightarrow Q$. Now, A and C are logically equivalent because both are contrapositive to each other and So, if C is true then A will also be true. So, for C, $(P \rightarrow Q) \rightarrow (P \rightarrow Q) \equiv X \rightarrow X \equiv True$ where $X \equiv P \rightarrow Q$ and for B, $(P \rightarrow Q) \rightarrow (Q \rightarrow P) \equiv \neg (\neg P \vee Q) \vee (\neg Q \vee P)) \equiv (P \wedge \neg Q) \vee (\neg Q \vee P) \equiv ((P \wedge \neg Q) \vee (\neg Q)) \vee P \equiv P \vee \neg Q$ and for D, $\neg(\neg P \vee Q) \vee (\neg P \wedge Q) \equiv (P \wedge \neg Q) \vee (\neg P \wedge Q) \equiv P \oplus Q$