Theorem pm2.61da2ne

Description: Deduction eliminating two inequalities in an antecedent. (Contributed by NM, 29-May-2013)

	Ref	Expression
Hypotheses	pm2.61da2ne.1	|- ( ( ph /\ A = B ) -> ps )
	pm2.61da2ne.2	|- ( ( ph /\ C = D ) -> ps )
	pm2.61da2ne.3	|- ( ( ph /\ ( A =/= B /\ C =/= D ) ) -> ps )
Assertion	pm2.61da2ne	|- ( ph -> ps )

Proof

Step	Hyp	Ref	Expression
1		pm2.61da2ne.1	|- ( ( ph /\ A = B ) -> ps )
2		pm2.61da2ne.2	|- ( ( ph /\ C = D ) -> ps )
3		pm2.61da2ne.3	|- ( ( ph /\ ( A =/= B /\ C =/= D ) ) -> ps )
4	2	adantlr	|- ( ( ( ph /\ A =/= B ) /\ C = D ) -> ps )
5	3	anassrs	|- ( ( ( ph /\ A =/= B ) /\ C =/= D ) -> ps )
6	4 5	pm2.61dane	|- ( ( ph /\ A =/= B ) -> ps )
7	1 6	pm2.61dane	|- ( ph -> ps )