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

Question

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?

Answer 1

$\lnot B\implies B$ is true when $B$ is true.

Answer 2

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 $((¬B ∧ (A \implies B)) \implies ¬(¬B\implies A))$