Polynomial time reduction from a language not in $\mathcal{NP}$

Question

Is it true that a language $L \notin \mathcal{NP}$ can not be reduced (in polynomial time) to a language in $Q \in \mathcal{NP}$ ? My take is that because the reduction has to be polynomial in time, i can't take in input a non NP problem, but i can't formalize my thoughts.

The question is part of a larger question where i have to prove that if $\mathcal{NP} \neq \mathcal{co-NP}$ then SAT is not reducible to UNSAT.

Answer 1

Assume $Q \in \mathcal{NP}$ (so there is non-determenistic Turing machine $T$ that solves it in polynomial time) and $L$ is polynomial reducible to $Q$. Then there is a non-determenistic machine $T'$ (defined as "apply reduction and launch $T$) that solves $L$ in (non-determenistic) polynomial time, thus $L \in \mathcal{NP}$.

Formally, the deduction is as follow:

1. $Q \in \mathcal{NP}$ (given)
2. $\neg(L \in \mathcal{NP})$ (given)
3. $Q \in \mathcal{NP} \rightarrow ((L \leq_P Q) \rightarrow (L \in \mathcal{NP})$ (proven above, $\leq_P$ means polinomaly reducible)
4. $(L \leq_P Q) \rightarrow (L \in \mathcal{NP})$ (m.p. from 1 and 3)
5. $\neg(L \in \mathcal{NP}) \rightarrow \neg (L \leq_P Q)$ (contraposition)
6. $\neg (L \leq_P Q)$ (m.p. from 2 and 5)