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 A compound sentence is a $\textit{tautology}$ if it is true independently of the truth values of its component atomic sentences. A sentence is $\textit{atomic}$ if it contains no sentential connectives.    
        
A sentence $P$ is said to $\textit{tautologically imply}$ a sentence $Q$ if and only if the conditional $P \rightarrow Q$ is a tautology.      
        
If $P$ and $Q$ are distinct atomic sentences, the sentence $\neg P \vee Q$ tautologically implies which of the following? (More than one option may be correct)     

  1. $\neg Q \rightarrow \neg P$       
          
  2. $Q \rightarrow P$       
           
  3. $P \rightarrow Q$       
          
  4. $\neg P \wedge Q$
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$P$ 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: 

  1. $\neg Q → \neg P$

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: $(\neg P \vee Q) → (\neg Q → \neg P) \equiv T → T \equiv T$

  1. $Q → P$

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: $(\neg P \vee Q) → (Q → P) \equiv T → T \equiv T$

  1. $P → Q$

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:  $(\neg P \vee Q) → (P → Q) \equiv T → T \equiv T$

  1. $\neg P \wedge Q$  

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, $\neg(\neg P \vee Q) \vee (\neg P \wedge Q) \equiv (P \wedge \neg Q) \vee (\neg P \wedge Q) \equiv P \oplus Q$

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