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3 Answers

Best answer
39 votes
39 votes
$$\begin{array}{|c|c|c|c|c|} \hline \textbf{p} & \textbf{q}& \textbf{p} \leftrightarrow \textbf{q} &\textbf{p} \to \lnot \textbf{q} & \begin{align*}(\boldsymbol{p} \leftrightarrow \boldsymbol{q}) & \to (\boldsymbol{p}\to \neg \boldsymbol{q}) \end{align*}\\\hline \text{0} & \text{0}& \text{1} & \text{1} & \text{1}\\\hline \text{0} & \text{1}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{1}& \text{1} & \text{0} & \text{0} \\\hline \end{array}$$
So, "imply" is FALSE making does not imply TRUE.
edited by
1 votes
1 votes
For implication, ( P ↔ Q ) → (P → Q’ ) should be tautology.

Checking with the by-case method,

Case1: P=T

            Q → Q’ = Q’

Case2: P=F

           Q’ → T = T.

From the case 1, we can conclude that the propostional expression isn’t a tautology.

So, we can say ( P ↔ Q ) doesn’t implies (P → Q’ ).

Hence, Given statement is TRUE.
edited by
–2 votes
–2 votes
by the given truth table in selected ans , does not imply is contingency, therefore its opposite(  imply) will also become contigency.    and the answer will be false
Answer:

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