# Is proof by contradiction “same thing” as $A \rightarrow B$ is true when $A$ is false?

I encountered earlier today a question "Is the proof by contradiction same as that $A \rightarrow B$ is true when $A$ is false?" continued by "Are they related, then? How?"

I think the answer is "no, they´re not the same thing", and demonstrates some confusion on the subject. However, as I am still more than a little confused about mathematical logic myself, too, I couldn't explain properly why this felt wrong to me. I pointed that proving statement $P$ by contradiction can be formulated symbolically as a $\neg P \rightarrow Q \wedge \neg Q$, and if $P = A \rightarrow B$ (because statements like that were the topic of discussion), we get something ...quite different. (I think I lost my trail of thought here.) However, it (the method of proof by contradiction) does feel like a different thing than the existence of a line in the truth table for $A \rightarrow B$ where $A$ is false and $B$ is true and thus the implication is true. Was I correct in this?

However, afterwards I remembered that $A \rightarrow B = \neg B \rightarrow \neg A$, where the right side is known as proof by contraposition. And the proof by contraposition is "in practice" usually done by assuming that $\neg B$ is true and then showing that $\neg A$ follows. (Because of the equality above, this is same as showing that "$A \rightarrow B$".)

Drawing the truth tables for both $A \rightarrow B$ and $\neg B \rightarrow \neg A$, this line (where $\neg B$ and $\neg A$ are true) does coincide with with $A$ being false and $B$ being true. So maybe one could say that $A \rightarrow B$ is true when $A$ is false is related to (or even "same thing as"?) the proof by contraposition.

Maybe we're both (or just me) confused about the relation of truth tables of statements like $A\rightarrow B$ to actually proving statements like "if A, then B".

A proof by contradiction is an argument of the form :

• assume $$A$$ and infer $$B$$;

• then, assume $$A$$ and infer $$\lnot B$$,

• and conclude with $$\lnot A$$.

In words, showing that the assumption leads to a contradiction, licences us to infer the negation of the assumption.

We cam express it in symbolic form (in propositional logic) as :

$$(A → B) → ((A → ¬B) → ¬A)$$.

We can easily reconcile the two versions noting that, by a "standard" result of a proof system for propositional logic (the Deduction Theorem), we have that :

if $$A \vdash B$$, then $$\vdash A \rightarrow B$$

where we have used (as usual) the "turnstile" ($$\vdash$$) to denote the relation of derivability in a proof system.

Thus, the above argument can be foramlized as follows :

if $$\vdash A \rightarrow B$$ and $$\vdash A \rightarrow \lnot B$$, then $$\vdash \lnot A$$.

How to reconcile it with truth table ?

Having $$\vdash A \rightarrow B$$ means to have proved the conditional $$A \rightarrow B$$, and the same with $$\vdash A \rightarrow \lnot B$$.

Of course we assume that "our logic" is sound, i.e. it does not prove false formulae.

Thus, if both formulae are true, due to the fact that one of $$B, \lnot B$$ is false, we must conclude that - in order that both conditionals are true - $$A$$ is false.

• What is an argument of the form "1. assume $\lnot$A, infer B. 2. assume $\lnot$A, infer $\not$B. 3. Conclude A"? Is that a proof by contradiction? – Doug Spoonwood Sep 15 '14 at 3:24
• @DougSpoonwood - yes, it is. – Mauro ALLEGRANZA Sep 15 '14 at 7:36

There is a connection between proving $A \rightarrow B$ and proving "if $A$, then $B$". That's called the deduction theorem, which guarantees that one is true iff the other is. You may want to look that up.

As for your question, I'm a little confused by your wording. A proof by contradiction works by showing, generally, that "if $A$, then $B$" is true, and then showing that $B$ is false, whence $A$ is false. Assuming the deduction theorem, this is the same as showing that $A \rightarrow B$ is true and $B$ is false. So yes, this is related to the idea that $A \rightarrow B$ is true when $A$ is false: given the truth of $A \rightarrow B$ and the falsity of $B$, the only way to jointly maintain this is to assume that $A$ is false.

In case you're unsure, draw the truth table, and then cross-out the row which say that $A \rightarrow B$ is false and the rows which say that $B$ is true. The remaining row will give you that $A$ is false.

• I think that's modus tollens or proof by contrapositive, not proof by contradiction. – WimC Sep 14 '14 at 19:36
• @WimC - They are connected. In a proof by contradiction, you assume the hypothesis and show it implies a contradiction. You can then apply modus tollens to negate the hypothesis. – Nagase Sep 14 '14 at 19:42

Suppose you want to prove statement $X$. If, assuming that $X$ is false leads to a contradiction of the form $Y\land \neg Y$, then, of course, $X$ must be true (by contradiction).

That is certainly the mainstream (i.e. non-instuitionst) view anyway.

• But by proving an implication using proof by contradiction we only prove that the implication is true but we don't know e.g. if the antecedent is true or false right? – user599310 Nov 1 '20 at 19:27
• There isn't necessarily any implication arising from a proof by contradiction. – Dan Christensen Nov 1 '20 at 23:14

A proof by contradiction consists of an argument where you hypothesize something, infer a conclusion which is a contradiction, and then discharge the hypothesis one way or another.

This means either that you infer either two propositions "A" and "$\lnot$A" which qualify as contradictory, or you infer a conjunction which has the form (A$\land$$\lnot$A), or more broadly you infer some formula which is known to consist of the negation of a tautology.