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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.

\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.
by

Could you please point out the flaw in my argument? 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?
 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?

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

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

okay :)

because at p = 1 and q = 1 the truth value of final expression is 0.
by looking truth table, it looks like LHS implies RHS.

because whenever LHS = T & RHS = F, at that time only it's giving F, rest of the cases it's giving T

how doesn't imply is true???
i think if we say that LHS implies RHS is TRUE  than for all the case it must be TRUE like a tautology but here we have contingency so we can say yes LHS doesn't implies RHS

but truth table obeys the truth table of implication truth table.

 p q $p\rightarrow q$ T T T T F F F T T F F T

from this table if we say that "p implies q" even when we can give a combination in which $p\rightarrow q$ is false.

like this, $(p\leftrightarrow q) \rightarrow (p\rightarrow \sim q)$ is false only when  $(p\leftrightarrow q)$ is T  &  $(p\rightarrow \sim q)$ is F.

rest of the cases it will true. It looks like an implication truth table.

see  if I say this  p⟺q implies p→¬q  and  i say it as true then for every combination of p and  q it should be true?

I think both the followings are different -

1. prove that $(p\Leftrightarrow q)\rightarrow (p\rightarrow \sim q)$ is a tautology.
2. prove that $(p\Leftrightarrow q)\rightarrow (p\rightarrow \sim q)$.

In 1st one if you show a combination of p,q such that LHS is T & RHS is F, you can declare that this is not a tautology.

In 2nd one you have to check whether LHS implies RHS or not, i.e you've to check whether this follows the implication rule(means truth table) or not.

Now according to the qsn. language, it feels like they want the 2nd one, not the 1st one.

@ can you tell me whether this statement is correct or wrong

"if I say this  p⟺q implies p→¬q  and  i say it as true then for every combination of p and  q it should be true"

correct
if you got false for some p and q then it means this statement is not correct on the other hand we can say it as   "

p⟺q does not imply p→¬q"
yes, you're correct. I got confused

@ so basically this question is asking whether (pq)→(p→∼q) is a tautology or not, isn't it?

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

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