978 views

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.

edited | 978 views

\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
+1

Could you please point out the flaw in my argument? 0
let, (p <--> q)= S  ,   (p --> not q) =R

from ur last second line

not S v Q =T..............i

not S v Q =F...............ii

from i and ii we can say

(not S v Q) v (not S V Q ) =T v F =T

NOT((not S v Q) v (not S V Q )) = NOT  ( T ) =F

can it be a prove?
+1
 P q p<->q p→¬q T T T F T F F T F T F T F F T T

Can you please explain  how p<->q does not imply p¬q is false?

+1
Sorry. "does not imply"  must be true. I have corrected.
0

It's ok. In one text book also they givep<->q does not imply p¬q is false .

+1
okay :)
0

because at p = 1 and q = 1 the truth value of final expression is 0.
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
how can I represent  " P does not imply Q"?
0
in Truth table have columns, for p, q, p double-implies q, p->~q
Now, check if third column implies the fourth.

1
2