Natural deduction Proof [(p →r) ∨ (q→r)]

Question

Trying to go from [(p →r) ∨ (q→r)] to prove (p∧q)→r.

Wanted to know if I am heading in the right direction with my deductions or where I am getting messed up.

1. (p→r) ∨ (q→r) premise
2. p assumption
3. p→r assumption
4. r →elim 2,3
5. q assumption
6. q→r assumption
7. r →elim 5,6
8. p∧q ∧intro 2,5
9. (p∧q)→r →intro 3-4,6-7,8

Answer 1

Wanted to know if I am heading in the right direction with my deductions or where I am getting messed up.

You are not. $p\wedge q$ should not be derived from assumptions of $p$ and $q$, it should be the assumption.

Always keep an eye on the goal.

You wish to prove $(p\wedge q)\to r$ from the premise $(p\to r)\vee (q\to r)$.

Therefore assume $p\wedge q$ aiming to derive $r$. To derive that, use a proof by cases.

$\begin{array}{|l}(p\to r)\vee(q\to r)\\\hline\begin{array}{|l} p\wedge q\\\hline\begin{array}{|l}p\to r\\\hline\vdots\\ r\end{array}\\\begin{array}{|l}q\to r\\\hline\vdots\\r \end{array}\\r\end{array}\\(p\wedge q)\to r\end{array}$