Discrete Mathematics | Propositional Logic | Test 1 | Question: 5

Question

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.    
      
When two sentences tautologically imply each other, they are said to be $\textit{tautologically equivalent.}$ P and Q are tautologically equivalent when and only when the biconditional $P \leftrightarrow Q$ is a tautology.       
       
If $P$ and $Q$ are distinct atomic sentences, the sentence $P$ tautologically equivalent to which of the following? (More than one option may be correct)     
        

• A. $\neg P \rightarrow P$      
         
• B. $P \rightarrow \neg P$      
         
• C. $P \vee Q$     
        
• D. $P \vee \neg P$

Answer 1

The question is asking :- Which all the options are (tautologically) equivalent to P?

• A. $P’ → P = (P’)’ + P = P$ and $P \leftrightarrow P$ is a tautology
• B. $P → P’ = P’ + P’ = P’$ and $P \leftrightarrow \neg P$ is not a tautology
• C. $P + Q$ and $P \leftrightarrow (P \vee Q)$ is not a tautology
• D. $P + P’ = True$ and $P \leftrightarrow True$ is not a tautology  

So, P is (tautologically) equivalent to option A only.