Contrapositive of $p\land q \implies r$ Now I understand that strictly speaking the contrapositive of this statement would be $\neg r \implies \neg p\lor\neg q$, but what I would like to do instead is prove that $\neg r\land q \implies \neg p$.
What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use?
Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?
The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.
 A: Hint
since $$\lnot r\Longrightarrow \lnot p\lor\lnot q$$then $$\lnot r\land q\Longrightarrow (\lnot p\lor\lnot q)\land q$$I leave to you to show that $$(\lnot p\lor\lnot q)\land q\equiv\lnot p$$
A: The contrapositive to $p \land q\implies r$ (which is what you are asking for in the title) is $\lnot r \implies \lnot(p \land q) = \lnot p \lor \lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p \implies q$).
This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:
Proposition: $x^4 - x^3 + x^2 \neq 1$, then $x \neq 1.$ 
The easy and elegant way to prove this is by contrapositive:
If $x = 1$, then $x^4 - x^3 + x^2 = 1$.
If you only think about what happens when $x^4 - x^3 + x^2 \neq 1$, it's not as trivial that $x \neq 1.$
There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.
