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I understand Implication, as follows:

p = rain stopped.
q = i go out.

p->q =  if "rain stopped" then "i go out".


p                       q                       p->q


F                       F                       T
(rain is not stopped)   (i don't go out)        (if rain is not stopped, I don't go out)
F                       T                       F
(rain is not stopped)   (i go out)              (if rain is not stopped, I go out)
T                       F                       F
(rain is stopped)       (i don't go out)        (if rain is stopped, I don't go out)
T                       T                       T
(rain is stopped)       (i go out)              (if rain is stopped, I go out)

But, this video says that $F \rightarrow T = T$

ryang
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user366312
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    Just because you go out, even when it is raining, does not signify that the statement $(p \implies q)$ is false. In fact, this situation is actually focusing on the critical difference between the statements $(p \implies q)$ and $(p \iff q)$. – user2661923 Aug 05 '21 at 10:28

2 Answers2

2

  1. The fact that the conditional $$P\to Q$$ is true whenever its antecedent $P$ is false (principle of explosion; vacuous truth) is actually so by definition:

          $P\to Q\,$ is a truth function that is tautologically equivalent to $\,\lnot P\lor Q.$

    So, $P\to Q\,$ is false precisely when $P$ is true but $Q$ false.

  2. To be clear: whenever $P$ is false, the assertion $P\implies Q\,$ gives no information about whether $Q$ is true.

  3. Summarising these two explanations of the motivation for the above definition:

    if we insist, to the contrary, that  False$\to$True  be false, then, unfortunately, these violations of natural deduction arise: $$\text{$A$ is true and $B$ is false $\implies\Big[(A\land B)\to A\Big]\;$is false!}$$ and $$\Big[\forall n\in\mathbb Z \;\big(n \text{ is a multiple of }4\, \to \,n \text{ is even}\big)\Big]\;\text{is false}!$$

  4. It is worth noting that in logic/mathematics, $P$ need not cause $Q$ for $P$ to imply $Q,$ that is, for the material conditional $\,P\to Q\,$ to actually be true.

    After all, the logical connective $\,\to,$ being a truth-functional operator, cares about truth states without considering the flow of time.

ryang
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0

Implication $P\Rightarrow Q$ is only false if $P$ is true and $Q$ is false (true premise and wrong conclusion). The video is right.

Wuestenfux
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