Give a natural deduction proof of $A \land (B \rightarrow \neg C)$ from hypotheses $A \land D$ and $A \rightarrow \neg C$.
My attempt so far:
- $A \land D$
- $A \rightarrow \neg C$
- A (1, $\land e$)
- $\neg C$ (2,3 $\rightarrow e$)
- D (1, $\land e$)
I'm new to the subject, how should I proceed from here?
Who cares about $D$?
You need to derive $B\to\neg C$, since you have already derive $A$, so you can use $\wedge$ introduction.
To derive a conditional, use conditional introduction. So derive $\neg C$ under the assumption of $B$.
Oh, look, you have already derived $\neg C$ without that assumption, so you can indeed do so under that assumption too.
You have actually done most of the work, so to finish up...
$${\begin{array}{|l}~~1.~A \land D\\~~2.~A \rightarrow \neg C\\\hline~~ 3.~A\qquad (1, \land e)\\~~~~\begin{array}{|l}~~4.~B\\\hline~~5.~\neg C\quad(2,3, \to e)\end{array}\\~~6.~B\to\neg C\quad (4{-}5,\to i)\\~~7.~A\wedge(B\to\neg C)\quad(3,6,\wedge i)\end{array}\\\blacksquare}$$