Is this implication true or false?

Question

When taking an if-then statement $P{\implies}Q,$ I'm confused about the Boolean logic, specifically when $P$ is false. I should have understood this much earlier, but for some reason it's just not clicking.

Take the statement (given that $x\in\mathbb R^+$) "if $x = 1,$ then $x^2 = 1.$" Clearly, if $x$ is not $1,$ then $x^2$ is not $1.$ Yet, how is it that in truth tables, if P is false, then the statement is seen as true? Where is my confusion?

No sorry @Nightflight, I'm assuming that x is only positive real numbers. — Daanyal Ali Akhtar, Jan 25 '22 at 01:08
You seem to be confusing the value of $P\rightarrow Q$ when $P$ is false vs. the value of $Q$ given $P\rightarrow Q$ and $\lnot P$. — Elliot Yu, Jan 25 '22 at 01:09
@ElliotYu Can you explain this further? Sorry. If possible, could you show me a truth table of the proposition I gave? Edit: Yes, it seems that the value of P => Q is true when P is false, correct? — Daanyal Ali Akhtar, Jan 25 '22 at 01:11
Also for the other case, In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? — JMoravitz, Jan 25 '22 at 01:37
@JMoravitz I don't think the OP is asking why F$\to$F should be (or makes sense to be, or is defined as) T per se; the duplicate targets are addressing this. Rather, I think the OP is (understandably) mixing up the truth value of a theorem's result $(Q)$ with the truth value of the theorem $(P{\implies}Q)$ itself. — ryang, Jan 25 '22 at 13:29

ryang · Answer 1 · 2022-01-26T16:54:40.253

2

You're conflating the implication $(P{\implies}Q)$'s truth value and the consequent $Q$'s truth value.

Given that $x=7,$ the statement $$x=1\implies x^2=1$$ is true, while the antecedent $$x=1$$ and the consequent $$x^2=1$$ are both false statements.

edited Jan 26 '22 at 16:54

answered Jan 25 '22 at 06:43

ryang

38,879
14
81
179

Is this implication true or false?

1 Answers1