Why aren't vacuous truths just undefined? [duplicate]

Solution 1:

Consider the statement:

All multiples of 4 are even.

You would say that statement is true, right?

So let's formulate that in formal logic language:

$\forall x: 4|x \implies 2|x$

(Here "$a|b$" means "$a$ divides $b$", that is, $b$ is a multiple of $a$.)

Now a $\forall$ statement is true if it is true whatever you insert for the quantified variable (after all,that's "for all" means). So let's try to insert $3$:

$4|3 \implies 2|3$

But wait, $4|3$ is false! Moreover, $2|3$ is also false. So the only way for the original statement to be true is that the implication $\text{false}\implies\text{false}$ gives true.

A similar argument can be done for $\text{false}\implies\text{true}$.

Solution 2:

This is done so that classical propositional calculus follows some natural rules. Let's try to motivate this, without getting into technical details:

The expression "$P\Rightarrow Q$" should be read "$P$ implies $Q$", or "whenever $P$ is true, $Q$ is also true".

The negation of such an expression would be a counter-example, i.e., "there is some case in which $P$ is true but $Q$ is not".

So assume $P$ is not true. The negation "$\lnot(P\Rightarrow Q)$" is not true in this case, by our interpretation above, so "$P\Rightarrow Q$" must be true.

We are basically using the rules that either an expression or its negation should be true, and that the negation of the negation of an statement is the statement itself. These are basic rules which are natural and useful, even though as a consequence we have that "$P\Rightarrow Q$" is true whenever $P$ is false.

Solution 3:

In a more informal sense, I like to think that $P\implies Q$ means that "$Q$ is at least as true as $P$". Which means that if $P$ is something false, then anything is "more true" than $P$, and thus the statement $P\implies Q$ is true.

Solution 4:

Consider any proof by contradiction, for example Euclid's proof of prime infinity.

If $x$ is the largest prime number ("P"), then $y=\prod_{p\,\,prime} p+ 1$ is a prime ("Q").

This is a valid implication that can be shown with perfectly valid simple arguments. However, the premise is never satisfied. But within the proof, the reader may not yet see that it is ''nonsense'' at this step. You just follow the logic step by step. The fact that the premise is never satisfied does not make the logic (used to derive the claim above) less valid. So it is reasonable to assume that the claim above is True.