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

Question

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. — John Douma, Aug 01 '24 at 04:32
By the definition of the implication. Also, this is almost certainly a duplicate. — Malady, Aug 01 '24 at 04:34
@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. — John Douma, Aug 01 '24 at 04:40
I was taught it as the definition. That’s interesting though. @JohnDouma — Malady, Aug 01 '24 at 04:42
@Malady Many define it that way because they are equivalent. But, if we just define it that way, then 佐武五郎 is correct; it is merely a convention. — John Douma, Aug 01 '24 at 04:43
Besides the usual definition with truth values, there is the definition of the implication in a system of natural deduction, using introduction and elimination rules. Then one can prove that $P\implies Q$ entails and is entailed by $\top$, whenever $P$ entails $\bot$. If OP finds the deduction rules convincing, then the truth tables for logical operators can be derived from them. — Jackozee Hakkiuz, Aug 01 '24 at 07:16
It is a definition. It is justified intuitively as fallows : "If A , then B" only applies if $A$ holds. If $A$ does not hold , we do not draw any conclusion. To consider however the conclusion RIGHT , if $A$ is false is not the only possibility. — Peter, Aug 01 '24 at 11:24
In fact , an argument based on $A$ and the conclusion $A\implies B$ is NOT correct if $A$ is false. Only the conlusion is correct. The consequence of this is that a single contradiction makes a whole theory useless (principle of explosion). This could have easily be avoided. — Peter, Aug 01 '24 at 11:25

Dan Christensen · Answer 1 · 2024-08-01T13:38:34.613

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

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

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