direct proof of propositional logic $B \vdash \neg(B \rightarrow (A \& \neg B))$ I got a tutorial question that I cant solve it...
the argument is this : $B \vdash \neg(B \rightarrow  (A \& \neg B))$
I have to use direct proof...
any help would be appreciated.
 A: It helps to first put the conclusion in a form that's easier to prove:
\begin{array}{l}
& \neg (B \to (A \land \neg B)) & \text{ Given }\\
& \neg (\neg B \lor (A \land \neg B)) & \text{ Material Imp. }\\
& B \land \neg (A \land \neg B) & \text{ DeMorgan }\\
& B \land (\neg A \lor B) & \text{ DeMorgan }\\
\end{array}
Then it can be proved as follows:
\begin{array}{l}
& \{1\} & 1. & B & \text{ Prem. }\\
& \{1\} & 2. & \neg A \lor B & \text{ 1 $\lor$I }\\
& \{1\} & 3. & B \land (\neg A \lor B) & \text{ 1,2 $\land$I }\\
& \{1\} & 4. & B \land \neg (A \land \neg B) & \text{ 3 DM }\\
& \{1\} & 5. & \neg (\neg B \lor (A \land \neg B)) & \text{ 4 DM }\\
& \{1\} & 6. & \neg (B \to (A \land \neg B)) & \text{ 5 MI }\\
\end{array}
Here's the proof by natural deduction:
\begin{array}{l}
& \{1\} & 1. & B & \text{ Prem. }\\
& \{2\} & 2. & A \land \neg B & \text{ Assum. }\\
& \{2\} & 3. & \neg B & \text{ 3 $\land$E }\\
& - & 4. & (A \land \neg B) \to \neg B & \text{ 2,3 CI }\\
& \{1\} & 5. & \neg (A \land \neg B) & \text{ 1,4 MT }\\
& \{6\} & 6. & B \to (A \land \neg B) & \text{ Assum. }\\
& \{1,6\} & 7. & A \land \neg B & \text{ 1,6 MP }\\
& \{1\} & 8. & (B \to (A \land \neg B)) \to (A \land \neg B) & \text{ 6,7 CI }\\
& \{1\} & 9. & \neg (B \to (A \land \neg B)) & \text{ 5,8 MT }\\
\end{array}
A: Before proceeding with the interference I would like to mention one handy tautological equivalence $$A \land (A \lor (X)) \leftrightarrow A \lor (A \land (X)) \leftrightarrow A$$ , where $X$ can be any atomic or compound sentence which results in true or false.
$$
\begin{array}{l}
& 1. & B & \text {Given}\\
& 2. & B \land (B \lor \neg A) & \text {Using the tautological equivalence asserted above}\\
& 3. & \neg \neg B \land \neg \neg (B \lor \neg A) & \text {Double negation}\\
& 4. & \neg (\neg B \lor \neg (B \lor \neg A)) & \text {De Morgan's Law}\\
& 5. & \neg (\neg B \lor (\neg B \land A)) & \text {Applying De Morgan's Law to $\neg (B \lor \neg A)$}\\
& 6. &\neg (B \to (\neg B \land A)) & \text {Equivalence for Implication and Disjunction }
\end{array}$$
