in Mathematical Logic recategorized by
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20 votes
20 votes

Let $p$ and $q$ be propositions. Using only the Truth Table, decide whether 

  • $p \Longleftrightarrow q$ does not imply $p \to \lnot q$

is True or False.

in Mathematical Logic recategorized by
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2 Answers

36 votes
36 votes
Best answer
$$\begin{array}{|c|c|c|c|c|} \hline \textbf{p} & \textbf{q}& \textbf{p} \leftrightarrow \textbf{q} &\textbf{p} \to \lnot \textbf{q} & \begin{align*}(\boldsymbol{p} \leftrightarrow \boldsymbol{q}) & \to (\boldsymbol{p}\to \neg \boldsymbol{q}) \end{align*}\\\hline \text{0} & \text{0}& \text{1} & \text{1} & \text{1}\\\hline \text{0} & \text{1}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{0}& \text{0} & \text{1} & \text{1}\\\hline \text{1} & \text{1}& \text{1} & \text{0} & \text{0} \\\hline \end{array}$$
So, "imply" is FALSE making does not imply TRUE.
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4 Comments

yes, you're correct. I got confused
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 @ so basically this question is asking whether (pq)→(p→∼q) is a tautology or not, isn't it?

 

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