Example 4

Let's construct a proof of:

~(A & B)

~(~B v C)

~(A v C)

Remember that we are now using our Equivalence Rules together with our Rules of Inference. We need both sets of rules to tackle proofs like this one. It's best again to begin by looking at our conclusion. It's the negation of a disjunction. There is no obvious way in which it appears in the premises. We know that we have an equivalence rule which transforms negations of disjunctions and conjunctions of negations. So we know that the conclusion is equivalent to a conjunction of ~A and ~C. If we can get each of ~A and ~C, then we can conjoin the two to get our conclusion. And that's exactly what we'll do:

line number	wff	justification	line number	commentary
1.	~(A & B)	premise
2.	~(~B v C)	premise
3.	~~B & ~C	D.M.	2	This will get us one of the conjuncts we want.
4.	~C	simp.	3	This is one of the conjuncts we need.
5.	~A v ~B	D.M.	1
6.	~~B	simp.	3	Need this for the next operation.
7.	~A	d.s.	5,6	Now we have the other conjunct.
8.	~A & ~C	conj	4,7	Almost there!
9.	~(A v C)	D.M.	8	done!