Theorem pm2.61dane

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 30-Nov-2011)

	Ref	Expression
Hypotheses	pm2.61dane.1	|- ( ( ph /\ A = B ) -> ps )
	pm2.61dane.2	|- ( ( ph /\ A =/= B ) -> ps )
Assertion	pm2.61dane	|- ( ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		pm2.61dane.1	|- ( ( ph /\ A = B ) -> ps )
2		pm2.61dane.2	|- ( ( ph /\ A =/= B ) -> ps )
3	1	ex	|- ( ph -> ( A = B -> ps ) )
4	2	ex	|- ( ph -> ( A =/= B -> ps ) )
5	3 4	pm2.61dne	|- ( ph -> ps )