The number of choices for [ , , , ], such that [(p q) ((p q) (( p) q))] is a tautology, is…

Question

The number of choices for $$\Delta  \in \{  \wedge , \vee , \Rightarrow , \Leftrightarrow \} $$, such that

$$(p\Delta q) \Rightarrow ((p\Delta  \sim q) \vee (( \sim p)\Delta q))$$ is a tautology, is :

Accepted Answer

Let $x:(p \Delta q) \Rightarrow(p \Delta \sim q) \vee(\sim p \Delta q)$

Case-I

When $\Delta$ is same as $v$

Then $(p \Delta \sim q) \vee(\sim p \Delta q)$ becomes

$(p \vee \sim q) \vee(\sim p \vee q)$ which is always true, so $x$ becomes a tautology.

Case-II

When $\Delta$ is same as $\wedge$

Then $(p \wedge q) \Rightarrow(p \wedge \sim q) \vee(\sim p \wedge q)$

If $p \wedge q$ is $T$, then $(p \wedge \sim q) \vee(\sim p \wedge q)$ is $F$

so $x$ cannot be a tautology.

Case-III

When $\Delta$ is same as $\Rightarrow$

Then $(p \Rightarrow \sim q) \vee(\sim p \Rightarrow q)$ is same at $(\sim p \vee \sim q) \vee$ $(p \vee q)$, which is always true, so $x$ becomes a tautology.

Case-IV

When $\Delta$ is same as $\Leftrightarrow$

Then $(p \Leftrightarrow q) \Rightarrow(p \Leftrightarrow \sim q) \vee(\sim p \Leftrightarrow q)$

$p \Leftrightarrow q$ is true when $p$ and $q$ have same truth values, then $p \Leftrightarrow \sim q$ and $\sim p \Leftrightarrow q$ both are false. Hence $x$ cannot be a tautology.

So finally $x$ can be $\vee$ or $\Rightarrow$.

Solutions · 1