Proof completion problem: I can only use primitive rules of inference, and I have contradictory premises. Standard proof completion:


*

*~(p&q) A

*~(~p&q) A

*~(p&~q) A

*~(~p&~q) A

*SHOW r


Contradicting r and then showing a contradiction seems like the obvious plan of attack, but after that I'm lost. I can only use primitive rules of inference--for instance, going from ~(A&B) to ~Av~B, or from knowing AvB and ~B to A are off limits. 
My attempts so far have run into the following problem: Obviously the premises form a contradiction, but I can't crack open the negation of each premise without listing one (a premise) on a SHOW line and then negating it, but that isn't helpful, since I already know what I'm proving.
 A: Of course, it depends on "the primitive rules of inference" you are licensed to use.
I'll give you an hint with Natural Deduction.
You have to use two "ausiliary" assumptions : $p$ and $q$.
We need two sub-proofs:
i) By $\land$-introduction, form $p$ and $q$, derive : $p \land q$.
This contradicts 1) and thus you can derive $\lnot p$, discharging the 1st ausialiary assumption $p$.
Now with $\lnot p$ and $q$ derive $\lnot p \land q$, contradicting 2); thus, derive $\lnot q$, discharging the 2nd ausiliary assumption $q$.
Up to now, we have derived $\lnot q$ with no undischarged assumptions.
ii) In the same way, using again $p$ and $q$ and premises 1) and 3), we derive $\lnot p$, with no undischarged assumptions.
Finally, we join the two "threads" deriving $\lnot p \land \lnot q$ by $\land$-introduction.
This contradicts 4) and thus, by $\lnot$-elimination (i.e. : $\varphi, \lnot \varphi \vdash \bot$) followed by $\bot$-elimination (i.e. : $\bot \vdash \psi$) or by Ex falso quodlibet, you can conclude with : $r$.
