How to prove $p \vee q \vdash \neg(\neg p \wedge \neg q)$?

Question

I'm trying to complete a list of exercises but this question I still can not solve. Somebody could help me?

updated: this is my conclusion

1 - $p\vee q$ Premise
2 - p Assume
3 - $\neg p\wedge\neg q$ Assume
4 - $\neg p$ $\wedge$e1 of 3
5 - $\perp$ $\neg$e 2,4
6 - $\neg(\neg p\wedge\neg q)$ $\neg$i 3-5

7 - q Assume
8 - $\neg p\wedge\neg q$ Assume
9 - $\neg q$ $\wedge$e2 of 8
10 - $\perp$ $\neg$e 7,9
11 - $\neg(\neg p\wedge\neg q)$ $\neg$i 8-10
12 - conclude $\neg(\neg p\wedge\neg q)$ $\vee$e 1, 2-6, 7-11

*Sorry for the text organization, I'm still learning how to use correctly.

Answer 1

Assume $p$.

Deduce $\neg (\neg p \land \neg q)$.

Assume $q$.

Deduce $\neg (\neg p \land \neg q)$.

Conclude $p \lor q \vdash \neg (\neg p \land \neg q)$

Answer 2

I want to consider this from a more computational perspective:

1) You can expand $\neg P$ as $P\to\,⊥$. Here "not P" is taken to mean: "A reason to believe $P$ would lead to absurdness".

So let's consider

$$(P\lor Q)\to \neg \left(\neg P\land \neg Q\right)$$

which by 1) translates to

$$(P\lor Q)\to\left(\left((P\to\,⊥)\land (Q\to\,⊥)\right) \to\,⊥\right)$$

At this point a case analysis would suffice to proof the statement. If you have a reason to believe $P$, say, then if you also had a reason to believe that $P$ (as well as $Q$) leads to absurdness, you're indeed lead to absurdness. The other case is handled analogously. This proof can expressed as

$x\mapsto\langle p,q\rangle\mapsto\mathrm{if}\ (x:P),\ \mathrm{do}\ p(x);\ \mathrm{elseif}\ (x:Q),\ \mathrm{do}\ q(x)$

2) Anyway, we can rewrite the proposition further... You can translate $P\to(Q\to R)$ to $(P\land Q)\to R$. This says "If a reason to believe $P$ implies that a reason to believe $Q$ suffices to show $R$, then, equivalently, reasons to believe both $P$ and $Q$ already imply $R$."

By this, we can rewrite the above as

$$\left((P\lor Q)\land\left((P\to\,⊥)\land (Q\to\,⊥)\right)\right) \to\,⊥$$

which, to make it more readable, we write as

$$\left((P\to\,⊥)\land (Q\to\,⊥)\land (P\lor Q)\right) \to\,⊥$$

In words, the statement just says: "Believing that $P$ is absurd and also believing that $Q$ is absurd, while at the same time believing that either $P$ or $Q$ is true ... this is absurd."