$(A → B) → C, A ∧ B \vdash C$
1.$\hspace{1cm}(A → B) → C \hspace{1cm}$premise
2.$\hspace{1cm}A ∧ B \hspace{2.5cm}$ premise
$\hspace{2cm}$ 3. $\hspace{1cm} A \to B \hspace{1cm}$ Assumption
$\hspace{2cm}$ 4. $\hspace{1cm} A \hspace{2cm} ∧ e_1 \hspace{0.5cm} 2$
$\hspace{2cm}$ 5. $\hspace{1cm} B \hspace{2cm} \to e \hspace{0.5cm} 3 $
6.$\hspace{1.75cm} C \hspace{2.5cm} \to e \hspace{0.5cm} 3, 1 $
I know this is wrong and I have tried multiple things but all of them feel wrong. My problem is that we already have A and B are true so I can't derive B by assuming A because of both are true. If I can't do that then how am I supposed to get C from the first premise. Also given A and B are true, I can't say that $A \to B$ , can I?
You can derive the conditional $A \to B$ by a technique Paul Tomassi refers to as augmentation on p. $64$ of his book Logic. First, place the second conjunct $B$ of the formula $A \wedge B$ on a line by itself via conjunction elimination (see line $3$ below). Next, assume the antecedent $A$ of the conditional $A \to B$ you want to derive (line $4$ below). Then, use conjunction introduction and conjunction elimination to derive the formula $A \wedge B$ yet again and once more place $B$ on a line by itself (lines $5$ and $6$ below). In doing so, you construct a subproof in which $B$ (on line $6$) follows from the assumption of $A$ (on line $4$), so you can now discharge the assumption and derive the conditional $A \to B$ via arrow introduction (line $7$ below).
$ \begin{array}{llll} \{1\} & 1. & (A \to B) \to C & \text{premise} \\ \{2\} & 2. & A \wedge B & \text{premise} \\ \{2\} & 3. & B & \text{2 $\wedge e_2$} \\ \{4\} & 4. & A & \text{Assumption} \\ \{2,4\} & 5. & A \wedge B & \text{3,4 $\wedge i$} \\ \{2,4\} & 6. & B & \text{5 $\wedge e_2$} \\ \{2\} & 7. & A \to B & \text{4,6 $\to i$} \\ \{1,2\} & 8. & C & \text{1,7 $\to e$} & \square \\ \end{array} $
Once you have the conditional $A \to B$, the remainder of the proof in which you derive $C$ is rather straightforward (see line $8$).