# How is it possible that if $A \implies B$ is true then $\lnot ( \lnot B \implies A )$ can be false?

While I was playing around with the material implication I made a proof by contradiction which I think it's wrong, but I don't find any mistake :

Say that $$A \implies B$$ is true , then suppose the truth of $$\lnot B \implies A$$ , but this can't be the case because otherwise $$\lnot B \implies A \implies B$$ , then $$\lnot( \lnot B \implies A)$$ is true .

However the truth table of the statement $$(A \implies B )\land \lnot( \lnot B \implies A)$$ isn't always true when $$A \implies B$$ is true , but I think it should be the case if my reasoning was correct. What's wrong with my proof?

• $A\implies B$ is equivalent to $\lnot B\implies\lnot A$ Jun 14, 2020 at 18:15

$$\lnot B\implies B$$ is true when $$B$$ is true.
Say that $$A \implies B$$ is true, then suppose the truth of $$\lnot B \implies A$$ , but this can't be the case because otherwise $$\lnot B \implies A \implies B$$ , then $$\lnot( \lnot B \implies A)$$ is true $$\dots$$ What's wrong with my proof?
$$A \implies B$$ does not implies $$\lnot B \implies A$$. I can see you want to construct a contradiction and here is what you said $$\lnot B \implies A \implies B$$ The idea is right. However, to make this reasoning work, we need to make sure $$\neg B$$ hold, therefore the correct conclusion should be $$(A\implies B)\implies\lnot( (\lnot B \implies A)\land\neg B)$$ Which is a tautology. Or we can say $$((¬B ∧ (A \implies B)) \implies ¬(¬B\implies A))$$