Theorem pm2.61da3ne 2597

Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013.)

Hypotheses

Ref	Expression
pm2.61da3ne.1	|- ((ph /\ A = B) -> ps)
pm2.61da3ne.2	|- ((ph /\ C = D) -> ps)
pm2.61da3ne.3	|- ((ph /\ E = F) -> ps)
pm2.61da3ne.4	|- ((ph /\ (A =/= B /\ C =/= D /\ E =/= F)) -> ps)

Assertion

Ref	Expression
pm2.61da3ne	|- (ph -> ps)

Proof of Theorem pm2.61da3ne

Step	Hyp	Ref	Expression
1		pm2.61da3ne.1	. 2 |- ((ph /\ A = B) -> ps)
2		pm2.61da3ne.2	. 2 |- ((ph /\ C = D) -> ps)
3		pm2.61da3ne.3	. . . 4 |- ((ph /\ E = F) -> ps)
4	3	adantlr 695	. . 3 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E = F) -> ps)
5		simpll 730	. . . 4 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E =/= F) -> ph)
6		simplrl 736	. . . 4 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E =/= F) -> A =/= B)
7		simplrr 737	. . . 4 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E =/= F) -> C =/= D)
8		simpr 447	. . . 4 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E =/= F) -> E =/= F)
9		pm2.61da3ne.4	. . . 4 |- ((ph /\ (A =/= B /\ C =/= D /\ E =/= F)) -> ps)
10	5, 6, 7, 8, 9	syl13anc 1184	. . 3 |- (((ph /\ (A =/= B /\ C =/= D)) /\ E =/= F) -> ps)
11	4, 10	pm2.61dane 2595	. 2 |- ((ph /\ (A =/= B /\ C =/= D)) -> ps)
12	1, 2, 11	pm2.61da2ne 2596	1 |- (ph -> ps)

Syntax hints:   -> wi 4   /\ wa 358   /\ w3a 934   = 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-3an 936  df-ne 2519

This theorem is referenced by: (None)