856 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 | 856 views

$p$ $q$ $p \leftrightarrow q$ $p \to \lnot q$ \begin{align*}(p \leftrightarrow q)&\to (p\to \neg q) \end{align*}
$0$ $0$ $1$ $1$ $1$
$0$ $1$ $0$ $1$ $1$
$1$ $0$ $0$ $1$ $1$
$1$ $1$ $1$ $0$ $0$

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

edited by
+1

Could you please point out the flaw in my argument?

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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 :)
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because at p = 1 and q = 1 the truth value of final expression is 0.
–1 vote
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"?
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in Truth table have columns, for p, q, p double-implies q, p->~q
Now, check if third column implies the fourth.

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