Fitch-Style Proof ¬(A → B) → (A ∧ ¬B) Good Evening,
I want to proof this statement: ¬(A → B) → (A ∧ ¬B). I have no premises.
I already did other proofs where I have no premises and a implication. I guess I have to start with the antecedent as my premise and try to generate the consequence. But there is not much I can use, so I thought I can proof it with contradiction.
This is my idea:

I used this website: http://proofs.openlogicproject.org/
HINT:
Your proof by contradiction is a good idea, but instead of doing that on the whole $A \land \neg B$ statement, try and prove each of $A$ and $\neg B$ using proof by contradiction.  It's clear how to set that up for the latter (assume $B$, and try and get $\bot$). For the former, you would assume $\neg A$, and if that leads to $\bot$, you get $\neg \neg A$, from which you can then infer $A$.
Observe this fact: $\neg(p\to q)\equiv p \wedge \neg q$ this is true because both propositions have the same truth table.
Now notice that $p\to p$ is a tautology.
Then use the fact, if $P\equiv Q$ then we can &quot;replace&quot; $P$ instead of $Q$ in any statement that includes $Q$.
Now $p\to p$ is a tautology then $\neg(p\to q)\to \neg(p\to q)$ is a tautology, and now
$\neg(p\to q)\to (p \wedge \neg q)$  is a tautology.