Need help solving a problem using the rules of inference - Discrete Math I am having issues solving this problem and I believe I would benefit if someone could help me break it down into simpler terms to perhaps give me a starting point. I tried applying DeMorgan's Law to the first line but I don't know what to do next or if that's even right.
$\exists x(A(x) \wedge \neg B(x))$
$\forall x(A(x) \rightarrow C(x))$

$\therefore \exists x(C(x)  \wedge \neg B(x))$
 A: $
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\oi#1{\qquad\mathbf{\lor I} \: #1 \\}
\def\oe#1{\qquad\mathbf{\lor E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\neg I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
The first premise says that something satisfies $A(x)$ and $\neg B(x)$. So, we can start making a substitution instance for it (heading towards the use of Existential Elimination). In my case, is a. Then, the second premise says that everthing satisfies $A(x) \to C(x)$. We can use Universal Elimination and choose that same name a. That decision enables the use of the other rules of inference.
As a next step, you will need to provide proofs of $C(a)$ and $\neg B(a)$.
$
\fitch{
1.\, \exists x(A(x) \land \neg B(x))\\
2.\, \forall x(A(x) \to C(x))
}{
  \fitch{3.\, A(a) \land \neg B(a)}{
   4.\,A(a) \to C(a) \Ae{2}
   \vdots\\
   k.\,C(a)\\
   \vdots\\
   l.\, \neg B(a)\\
   \vdots
}\\
m. \exists x(C(x) \land \neg B(x))
}
$
Can you see how to fill the blanks ?
A: *

*$\exists x(A(x) \wedge \neg B(x))$ premise

*$\forall x(A(x) \rightarrow C(x))$ premise

*$A(k) \wedge \neg B(k)$ by existential instantiation, 1

*$A(k) \rightarrow C(k)$ by universal instantiation, 2

*$A(k)$ by conjunction elimination, 3

*$C(k)$ by modus ponens, 4,5

*$\neg B(k) \wedge A(k)$ by commutativity, 3

*$\neg B(k)$ by conjunction elimination, 7

*$C(k) \wedge \neg B(k)$ by conjunction introduction, 6,8

*$\exists x(C(x)  \wedge \neg B(x))$ by existential generalization, 9
