yes, you're correct. I got confused

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$$\begin{array}{|c|c|c|c|c|} \hline \textbf{p} & \textbf{q}& \textbf{p} \leftrightarrow \textbf{q} &\textbf{p} \to \lnot \textbf{q} & \begin{align*}(\boldsymbol{p} \leftrightarrow \boldsymbol{q}) & \to (\boldsymbol{p}\to \neg \boldsymbol{q}) \end{align*}\\\hline \text{0} & \text{0}& \text{1} & \text{1} & \text{1}\\\hline \text{0} & \text{1}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{1}& \text{1} & \text{0} & \text{0} \\\hline \end{array}$$

So, "imply" is FALSE making does not imply TRUE.

So, "imply" is FALSE making does not imply TRUE.

@MRINMOY_HALDER so basically this question is asking whether (*p*⇔*q*)→(*p*→∼*q*) is a tautology or not, isn't it?

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edited
Mar 25, 2021
by subbus

In last column of above truth table $p \Longleftrightarrow q \rightarrow p \to \lnot q$ doesn’t have all 1’s.Therefore, it is not a tautology

If It is a tautology(all 1’s in last column) then $p \Longleftrightarrow q$ implies $p \to \lnot q$

i.e., $ \left( p \Longleftrightarrow q~~~ \rightarrow ~~~ p \to \lnot q \right) \equiv True $

But Now it is not tautology so $p \Longleftrightarrow q$ does not implies $p \to \lnot q$

ie. $\left( p \Longleftrightarrow q ~~~ \not \rightarrow ~~~p \to \lnot q \right) \equiv True$

If It is a tautology(all 1’s in last column) then $p \Longleftrightarrow q$ implies $p \to \lnot q$

i.e., $ \left( p \Longleftrightarrow q~~~ \rightarrow ~~~ p \to \lnot q \right) \equiv True $

But Now it is not tautology so $p \Longleftrightarrow q$ does not implies $p \to \lnot q$

ie. $\left( p \Longleftrightarrow q ~~~ \not \rightarrow ~~~p \to \lnot q \right) \equiv True$

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