Proof logical statement with interference rules

Question

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

1. $B$ (premise)
2. $B \rightarrow D$ (premise)
3. $D$ (Modus ponens 1,2)
4. $A \rightarrow C $ (premise)
5. $\neg C$ (premise)
6. $\neg A$ (Modus tollens 5,4)
7. $\neg A \wedge D$ (Conjunction introduction 6, 3)

   ---

   Q.E.D

Is my proof correct?

Answer 1

You really have only one premise:

$$ \neg C \wedge B \wedge (A \rightarrow C) \wedge (B \rightarrow D )$$

Thus, you'll need to infer the statements that you have on lines 1,2,4, and 5 from this premise using the Simp rule (this is short for Simplification ... other textbooks call this conjunction elimination)

Also, for line 7, the name of the rule in your system is Conj (short for Conjunction) ... you call this conjunction introduction, and indeed many textbooks do, but in your system it is Conj

Otherwise, your proof looks to be fine!