Proof following statement with interference rules ( without truth table) that
$$ (\neg C \wedge B \wedge (A \rightarrow C) \wedge (B \rightarrow D ) )\implies (\neg A \wedge D ) $$
Attempt to proof
- $B$ (premise)
- $B \rightarrow D$ (premise)
- $D$ (Modus ponens 1,2)
- $A \rightarrow C $ (premise)
- $\neg C$ (premise)
- $\neg A$ (Modus tollens 5,4)
- $\neg A \wedge D$ (Conjunction introduction 6, 3)
Q.E.D
Is my proof correct?