Theorem pm2.61da2ne 2596

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

Hypotheses

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

Assertion

Ref	Expression
pm2.61da2ne	|- (ph -> ps)

Proof of Theorem pm2.61da2ne

Step	Hyp	Ref	Expression
1		pm2.61da2ne.1	. 2 |- ((ph /\ A = B) -> ps)
2		pm2.61da2ne.2	. . . 4 |- ((ph /\ C = D) -> ps)
3	2	adantlr 695	. . 3 |- (((ph /\ A =/= B) /\ C = D) -> ps)
4		pm2.61da2ne.3	. . . 4 |- ((ph /\ (A =/= B /\ C =/= D)) -> ps)
5	4	anassrs 629	. . 3 |- (((ph /\ A =/= B) /\ C =/= D) -> ps)
6	3, 5	pm2.61dane 2595	. 2 |- ((ph /\ A =/= B) -> ps)
7	1, 6	pm2.61dane 2595	1 |- (ph -> ps)

Syntax hints:   -> wi 4   /\ wa 358   = wceq 1642   =/= wne 2517

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-an 360  df-ne 2519

This theorem is referenced by:  pm2.61da3ne  2597