# False implies anything [duplicate]

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 $$F \rightarrow T = T$$

• 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 $(p \iff q)$. Aug 5 '21 at 10:28

## 2 Answers

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

$$P\to Q\,$$ is the 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: when $$P$$ is false, $$\,P\to Q\,$$ (is always true so) never gives any information about $$Q.$$

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: $$\text{A is true and B is false \implies\bigg(\big[(A\land B)\to A\big]\;is false}\bigg)$$ 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 $$Q$$ (i.e., for the material conditional $$P\to Q$$ to actually be true); after all, $$\to$$ is defined truth-functionally rather than based on interpretation (of the meanings of $$P$$ and $$Q$$).

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