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


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.

  • $\begingroup$ 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

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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