I'm doing proof by contradiction for the first time and I understand the concept now, but I can't find a contradiction in my negation.
I'm trying to conclude $\neg n$
$h\wedge\neg r$
$(h\wedge n)\to r$
$~n$
My negations are:
$\neg h\vee r$
$h\wedge n\wedge\neg r$
Remember, proof by contradiction works by assuming the negation of your conclusion, and inferring a contradiction - it has nothing to do with negating your premises.
Think about the problem in terms of the intuitive logic of the situation. Your premises say the following: first, "$h$ is true, and $r$ is not"; second, "if $h$ and $n$ are both true, then $r$ is true". Intuitively, if $n$ were true (the negation of your conclusion), we would have that $h$ and $n$ are both true (using the first half of the first premise). So $r$ is true (using the second premise). But $r$ is false (using the second half of the first premise), which means we now have $r \wedge \neg r$, a contradiction.
Now, the challenge is just taking those steps and converting them into formal logic.
The negations of the two premises are not useful. You need to negate the sentence you are trying to prove, but the premises you use as they are, like this:
Assume $n$. From $h\wedge\neg r$ follows $h$. Infer $h\wedge n$. Use modus ponens to infer $r$. From $h\wedge\neg r$ follows also $\neg r$. Contradiction. Conclude $\neg n$.
$h \land \neg r \qquad \qquad$ Premise
$ (h \land n) \rightarrow r \qquad$ Premise
$\qquad n \qquad \qquad \qquad$ Assumption (for the proof by contradiction)
$\qquad h \qquad \qquad \qquad$ Simplification 1
$\qquad h \land n\qquad \qquad$. Conjunction 3,4
$\qquad r \qquad \qquad \qquad$ Modus Ponens 2,5
$\qquad \neg r \qquad \qquad$ Simplification 1
$\qquad \bot \qquad \qquad$ Contradiction 6,7
$\neg n \qquad \qquad$ Proof by Contradiction 3-8
$$h\land \neg r$$ $$(h\land n) \rightarrow r$$
$$(h\land \neg r)\rightarrow\neg r$$ $$(h\land \neg r)\rightarrow h$$ $$\neg r \rightarrow \neg(h\land n)$$ $$(\neg(h\land n)\land h)\rightarrow\neg n$$