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+13 votes
756 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.

asked in Mathematical Logic by Veteran (59.6k points)
edited by | 756 views

3 Answers

+23 votes
Best answer
$p$ $q$ $p \leftrightarrow q$ $p \to \lnot q$ $\begin{align*}(p \leftrightarrow q)&\to (p\to \neg q) \end{align*}$
$0$ $0$ $1$ $1$ $1$
$0$ $1$ $0$ $1$ $1$
$1$ $0$ $0$ $1$ $1$
$1$ $1$ $1$ $0$ $0$

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

answered by Veteran (363k points)
edited by
+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
So, answer is true right?

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
answered by Junior (739 points)
–2 votes
how can I represent  " P does not imply Q"?
answered by Active (2.6k points)
0
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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