# Let $P,Q$ be statements. If $P$ is a false statement, then $P\implies Q$ is true. This convention works perfectly anytime. Why? [duplicate]

I learned the following convention.

Convention 1:

Let $$P,Q$$ be statements.
If $$P$$ is a false statement, then $$P\implies Q$$ is true.

Convention 1 works perfectly anytime.
Why?

For example,

Proposition 1:

Suppose $$n\in\mathbb{N}$$.
Then for any $$x\in n$$, $$x\in\mathbb{N}$$.

Proof:

If $$n=0$$, then by Convention 1, for any $$x\in 0,x\in\mathbb{N}$$.

Suppose $$n\in\mathbb{N}$$ and for any $$x\in n$$, $$x\in\mathbb{N}$$.

Let $$x$$ be any element of $$n\cup\{n\}$$.
If $$x\in n$$, then $$x\in\mathbb{N}$$ by induction hypothesis.
If $$x=n$$, then $$x=n\in\mathbb{N}$$.

So, by induction, if $$n\in\mathbb{N}$$, then for any $$x\in n$$, $$x\in\mathbb{N}$$.

I think the above proof is the standard proof but I would like to prove Proposition 1 as follows and tend to prove Propostion 1 as follows:

Proof:

If $$n=0$$, then by Convention 1, for any $$x\in 0,x\in\mathbb{N}$$.

Suppose $$n\in\mathbb{N}$$ and for any $$x\in n$$, $$x\in\mathbb{N}$$.

Let $$x$$ be any element of $$n\cup\{n\}$$.
If $$n=0$$, then $$x=n$$ since $$x\notin n$$.
And $$x=n\in\mathbb{N}$$.
Suppose $$n\neq 0$$.
Let $$x$$ be any element of $$n\cup\{n\}$$.
If $$x\in n$$, then $$x\in\mathbb{N}$$ by induction hypothesis.
If $$x=n$$, then $$x=n\in\mathbb{N}$$.

So, by induction, if $$n\in\mathbb{N}$$, then for any $$x\in n$$, $$x\in\mathbb{N}$$.

Convention 1 works perfectly anytime.
Why?

• It's not a convention. We can show that $P\implies Q$ is equivalent to $\lnot P\lor Q$. Since only one of the elements of a disjunction must be true and $\lnot P$ is true when $P$ is false, the result follows. Commented Aug 1 at 4:32
• "Ex falso quodlibet" Commented Aug 1 at 4:33
• By the definition of the implication. Also, this is almost certainly a duplicate. Commented Aug 1 at 4:34
• @Malady That's actually not by definition. $P\implies Q$ simply says that $Q$ is true whenever $P$ is true. We also know that $\lnot P\lor P$ is always true so if $P\implies Q$ is true, we get $\lnot P\lor Q$ by substitution. If we assume $\lnot P\lor Q$ is true, then if $P$ is true, $Q$ must be true because $\lnot P$ is false. Therefore, we get $(P\implies Q)\implies (\lnot P\lor Q)$ and vice versa, so the two statements are equivalent. Commented Aug 1 at 4:40
• I was taught it as the definition. That’s interesting though. @JohnDouma Commented Aug 1 at 4:42

It seems "counter-intuitive" if you mistakenly think that "A implies B" means "A causes B." Or that B is always true when A is true. In classical propositional logic, "A implies B" simply rules out only the possibility that both A is true and B is false. This is entirely consistent with the usual truth table for "A implies B":

In this table, we see that where both $$A$$ and $$A \implies B$$ are true (on line 1 only), $$B$$ must also be true, thus implementing the detachment rule.

Note, too, that where A is false (on lines 3 and 4), we have $$A \implies B$$ being be true regardless of the truth value of $$B$$, thus implementing the principle of vacuous truth. This form of argument is rarely, if ever, used in daily discourse as we rarely, if ever, consider the implications of propositions known to be false. It is, however, routinely used in very technical arguments, e.g. in mathematical proofs.

Text version of truth table:

A B A=>B

T T T

T F F

F T T

F F T