Proving ${\sim p}\mid{\sim q}$ implies ${\sim}(p \mathbin\& q)$ using Fitch I am struggling with proving something in Fitch. How can I prove from the premise ~p | ~q , that ~(p & q). Any ideas on how I should proceed?
 A: I'll have my first stab at an answer here (I've been won over from Stack overflow- first answer!).
I believe on math.stackexchange it's not tradition to completely give the game away if it's a homework question?
notation: noticed you're using single bar for vel, I'll use $\vee$ and "|" to distinguish Fitch hypothesis marks until I can work out math stackexchange formatting.
So,
Hint: your last step will be $¬(p \wedge q)$, of course, and obviously your first
1 $(¬p \vee ¬q)$
Then we prove by two classical reductio.
First, on the conclusion
2 | $p \wedge q$
Which gives us
3 | $p$
Then, remember the $\vee$-elim rules (you're working from (1) now): so we assume:
4 | | $¬p$
5 | | $\bot$(by 3)
Well, I'll leave you to fill in the details, but we can do the same for the q side of 2 and 1.
This means we can $\vee$ eliminate to
9) | $\bot$
Which gives us the discharge by reductio of (2)
10) $¬(p \wedge q)$
Hope this helps you fill in the minimal gaps.
If you wanted to cheat you could have a look at de morgan's laws as you're going one way here, or ask whether you can prove this in all logics.
Apologies for my first attempt at formatting, I'll try to edit as I go
A: I came up with the same derivation (see Tom's answer), but it's in a different format, so might be of some tiny use to the OP:

