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I get confused when i think about logical equivalence between conditional statements. For example saying that

∼(P⇒Q)=P∧∼Q.

If there is variables involved then the statement on the left says that there exists some value of that variable for which P does not imply Q. The statement on the right P and not Q is true for all values of that variable, This is what i understand but i think im wrong because these statements are logically equivalent and are meant to say the same thing about P and Q.

I can understand if they did not include variables but not if they do can someone help explain, thanks.

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  • $\begingroup$ Equivalence here simply means that the two statements have the same truth table, entry by entry. For any assignment of truth values for $P,Q$, the two statements are either both true ir both false. $\endgroup$ – AnyAD Dec 1 '18 at 23:59
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The way you are trying to understand equivalence is slightly correct, but mostly not: first one is saying $P$ does not imply $Q$. In a truth-table, if you notice, a conditional is true in two and only two cases: either $P$ is false or $P$ is true and $Q$ is true. Therefore, to say $P$ does not imply $Q$ is to say: $P$ is true and $Q$ is false.

Now notice, $P$ and not $Q$ is true exactly when both $P$ is true and $Q$ is false. Did you notice anything similar between this sentence and the last sentence of the first paragraph?

This similarity means $\lnot (P \implies Q)$ and $P \land \lnot Q$ have the same truth-table. Therefore, they are uquivalent.

That said, falisty and truth of a propositional statement depends on the variables with respect to the assignment of truth values to those variables.

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    $\begingroup$ Thanks i understand This in terms of statements which do not involve variables but my point is when you have P implies Q and P and Q involve variables yes this statement is only false when P is true and Q is false but there may be some values of the variable which make P true and Q true and therefore the statement P and not Q would be false $\endgroup$ – Carlos Bacca Dec 2 '18 at 0:01
  • $\begingroup$ @CarlosBacca I think you misunderstand a logical implication. To say $P$ implies $Q$ is to say if $P$ is true Then $Q$ must be, by definition true. In order for the conditional to makes sense. Therefore, you can not isolate propositions involved in a conditional like you have done. $P\implies Q$ is true when $P$ and $Q$ are semantically and syntactically related. Not isolated. $\endgroup$ – Bertrand Wittgenstein's Ghost Dec 2 '18 at 0:06
  • $\begingroup$ Thanks so say you have x is even implies x is a multiple of 6 which is sometimes true and sometimes false, so it is false. this statement is equivalent to x is even and x is not a multiple of 6 again sometimes true and sometimes false i dont get it $\endgroup$ – Carlos Bacca Dec 2 '18 at 0:11
  • $\begingroup$ @CarlosBacca x even implies x is a multiple of 6 is false though. Since 2 is even, but 2 is not a multiple of 6. Therefore, x is even implies x is a multiple of six is not by definition true. Just a rule of thumb, equivalences are determined by truth table. Using particular examples can not prove or disprove an equivalence. $\endgroup$ – Bertrand Wittgenstein's Ghost Dec 2 '18 at 0:36

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