Is it possible to eliminate a contradiction without recourse to the principle of explosion? I'd like to derive the following inference rule:

$$
\frac{p\lor(q\land\neg q)}{p}\quad\text{[ContradictionElimination]}
$$

I assumed that I could do this minimally somehow, however it turns out I need an alternative form of the principle of explosion. My derivation is:

Rule (ContradictionElimination)
  Premise
    P∨(Q∧⌐Q)
  Conclusion
    P
  Proof
    Suppose
      P
    Hence
      P
    P=&gt;P

    Suppose
      Q∧⌐Q
    Then
      Q
      ⌐Q
    Hence
      P by PrincipleOfExplosionAlternativeForm
    Q∧⌐Q=&gt;P

    P by DisjunctionElimination


My alternative form of the principle of explosion is, by the way:

$$
\frac{p\quad\neg p}{q}\quad\text{[PrincipleOfExplosionAlternativeForm]}
$$

This is easy enough to derive from the standard principle of explosion and modus ponens.

Without a way to eliminate contradictions minimally, so to speak, all my minimal proofs of De Morgan's laws become intuitionsitic. This seems wrong to me.
I'm going to try to answer this question although I'm not entirely sure the answer is right. Maybe it will encourage debate, though. I think that the answer is no, you cannot derive this rule without some form of the principle of explosion. My reasoning goes as follows...

The premise is a disjunction $p\lor(q\land\neg q)$. I have chosen to arrive at the conclusion by proof by cases, disjunction elimination in other words, which means I have to deal with $q\land\neg q$ somehow. The only other way that I can see to make progress from the premise is to use the rule that distributes disjunction over conjunction. This gives $(p\lor q)\land(p\lor\neg q)$. Now I have two disjunctions to deal with. Taking the rightmost $p\lor q$, I can only proceed by proof by cases again. Assuming $p$, I'm done. Assuming $q$ on the other hand, I must make use of the other side of the conjunction, namely $p\lor\neg q$ and I must then proceed to $q\land(p\lor\neg q)$. Using the rule that conjunction distributes over disjunction I get, unfortunately, $(p\land q)\lor(q\land\neg q)$ and I'm no better of than where I started.

My line of reasoning is based on there only really being one way that the derivation can proceed. However it seems exhaustive, so I think there's something in it. It also seems, from an intuitive standpoint, that you can't eliminate $q\land\neg q$ without something like the principle of explosion, although admittedly this intuition is tainted by my findings above. Perhaps some other interpretation would shed more light on this.
In Polish notation you might use the rule of inference

CN$\alpha$K$\beta$N$\beta$ $\vdash$ $\alpha$

which I'll call No.

I'll also use 

Kol: K$\alpha$$\beta$ $\vdash$ $\alpha$

Kor: K$\alpha$$\beta$ $\vdash$ $\beta$

Ki: $\alpha$, $\beta$ $\vdash$ K$\alpha$$\beta$

and

Ao: A$\alpha$$\beta$, C$\alpha$$\gamma$, C$\beta$$\gamma$ $\vdash$ $\gamma$

The proof then can go:

premise   1 ApKqNq
suppose   2 | p
Ci 2-2    3 Cpp
suppose   4 | KqNq
suppose   5 || Np
Kol 4     6 || q
Kor 4     7 || Nq
Ki 6 7    8 || KqNq
Ci 5-8    9 | CNpKqNq
No 9      10 | p
Ci 4-10   11 CKqNqp
Ao 1 3 11 12 p