$Option – D : p=T$ and $q=F$ will make it false. So, it is not necessarily tautology.

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Let $p,q$ be two atomic propositional assertions. Then which of the following is/are false?

- $(p \rightarrow q) \vee (p \rightarrow \neg q)$ is necessarily tautology.
- $(p \rightarrow q) \vee (q \rightarrow p)$ is necessarily tautology.
- $(p \rightarrow q) \vee (q \rightarrow \neg p)$ is necessarily tautology.
- $(p \rightarrow q) \vee (\neg q \rightarrow \neg p)$ is necessarily tautology.

1 vote

For Option A,

$\begin{array}{|c|c||c|c|c|c|} \hline p & q & p\rightarrow q & \neg q & p\rightarrow \neg q &(p\rightarrow q)\vee (p\rightarrow \neg q)\\\hline T & T & T & F& F & T\\\hline T & F & F& T & T & T \\\hline F & T & T & F & T & T\\\hline F & F & T & T & T & T \\ \hline\end{array}$

The formula is valid since it is satisfied by every interpretation. Similarly, we can check for other options.

$\begin{array}{|c|c||c|c|c|c|} \hline p & q & p\rightarrow q & \neg q & p\rightarrow \neg q &(p\rightarrow q)\vee (p\rightarrow \neg q)\\\hline T & T & T & F& F & T\\\hline T & F & F& T & T & T \\\hline F & T & T & F & T & T\\\hline F & F & T & T & T & T \\ \hline\end{array}$

The formula is valid since it is satisfied by every interpretation. Similarly, we can check for other options.