Natural Deduction Proof without Logical Equivalences Natural Deduction (you cannot use logical equivalences), prove the following:
$$\frac{(p\lor q)\land(r\rightarrow \lnot p), \space q \rightarrow \lnot r}{\therefore \lnot r}$$
So far I have the following:
$1. \space (p\lor q)\land(r\rightarrow \lnot p)\qquad$ Premise.
$2. \space q\rightarrow \lnot r \qquad \qquad \qquad $ Premise.
$\boxed{3. \space p \lor q \qquad \qquad \qquad Assumption. \\ 
4. \space r \rightarrow \lnot p \qquad \qquad \space \space \land-eliminatation(3
)}$
Now I don't know where to go. I am lost on what we are trying to prove. How are proposition $1$ and proposition $2$ linked? I know I want to obtain the last statement $\lnot r$, but don't know what to do after $r$.
Am I supposed to find out $p \lor q$ is true by showing $p$ is true, and if $p$ is true then for $r \rightarrow \lnot p$, $r$ has to be false? Then $q$ must be true then $\lnot r$ equals $T$ leading to $\lnot r$? Is this what I should be thinking?
 A: 
Am I supposed to find out $p \lor q$ is true ...?

As Mauro says in the comments, there is no need to assume $p \lor q$ is true. It is derived from one of your premises using $\land$-Elimination.
As you goal is $\lnot r$, and you can derive (no need to assume) $p \lor q$, if you can get $\lnot r$ assuming $p$ is true and obtain that same statement when assuming $q$, then, with the use of $\lor$-Elimination, you are allowed to write $\lnot r$.
A possible proof skeleton would be:
$
\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 \\}
$
$
\fitch{
(p\lor q) \lor (r \to \lnot p)\\
q \to \lnot r
}{
  p \lor q\\
  r \to \lnot p\\
  \fitch{p}{
   \fitch{r}{
     \vdots
}
}\\
\fitch{q}{
 \vdots
}\\
\lnot r
}
$
A: As correctly said by Mauro curto in his comment, the missing step in your attempt of derivation is the use of the inference rule $\lor \mathbf{E}$ for eliminating the disjunction $p \lor q$.
The idea is that, because of the first premise $(p \lor q) \land (r \to \lnot p)$, the disjunction $p \lor q$ holds but it is unknown if $p$ holds or $q$ holds. In the first case, since $r \to \lnot p$, you can easily infer $\lnot r$ (via modus tollens). In the second case, $\lnot r $ immediately follows because of the second premise.
Therefore, a correct derivation in natural deduction of $\lnot r$ from the premises $(p \lor q) \land (r \to \lnot p)$ and $q \to \lnot r$ is the following:
$
\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 \\}
$
$
\fitch{1. \, (p \lor q) \land (r \to \lnot p) \qquad \text{premise}
 \\
  2.\, q \to \lnot r \qquad \text{premise}
} 
{  3. \, p \lor q \ce{(1)}
 \fitch{4.\, p \qquad \text{assumption}}
    { 5. \, r \to \lnot p  \ce{(1)} 
\fitch{6. \, r \qquad \text{assumption}}
             {
             7. \, \lnot p  \ie{(6, 5)}
             8. \, \bot \ne{(7, 4)}
             }
\\
9. \, \lnot r \ni{(6{-}8)}
}\\
\fitch{10.\, q \qquad \text{assumption}}
    { 11. \, \lnot r \ie{(2, 10)}
}\\
 12. \, \lnot r \oe{(3{-}11)}
}
$

Note that in your attempt of derivation, $p \lor q$ need not be assumed, because it follows from the first premise $(p \lor q) \land (r \to \lnot p)$ by means of the inference rule $\land \mathbf{E}$ for elimination of conjunction.
