Theorem pm2.61da3ne 2883

Description: Deduction eliminating three inequalities in an antecedent. (Contributed by NM, 15-Jun-2013.) (Proof shortened by Wolf Lammen, 25-Nov-2019.)

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.2	. 2 |- ( ( ph /\ C = D ) -> ps )
2		pm2.61da3ne.3	. 2 |- ( ( ph /\ E = F ) -> ps )
3		pm2.61da3ne.1	. . . . 5 |- ( ( ph /\ A = B ) -> ps )
4	3	a1d 25	. . . 4 |- ( ( ph /\ A = B ) -> ( ( C =/= D /\ E =/= F ) -> ps ) )
5		pm2.61da3ne.4	. . . . . 6 |- ( ( ph /\ ( A =/= B /\ C =/= D /\ E =/= F ) ) -> ps )
6	5	3exp2 1285	. . . . 5 |- ( ph -> ( A =/= B -> ( C =/= D -> ( E =/= F -> ps ) ) ) )
7	6	imp4b 613	. . . 4 |- ( ( ph /\ A =/= B ) -> ( ( C =/= D /\ E =/= F ) -> ps ) )
8	4, 7	pm2.61dane 2881	. . 3 |- ( ph -> ( ( C =/= D /\ E =/= F ) -> ps ) )
9	8	imp 445	. 2 |- ( ( ph /\ ( C =/= D /\ E =/= F ) ) -> ps )
10	1, 2, 9	pm2.61da2ne 2882	1 |- ( ph -> ps )

Syntax hints:   -> wi 4   /\ wa 384   /\ w3a 1037   = wceq 1483   =/= wne 2794

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 197  df-an 386  df-3an 1039  df-ne 2795

This theorem is referenced by:  trljco  36028  dvh4dimN  36736