Theorem pm2.61ne 2592

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 24-May-2006.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Hypotheses

Ref	Expression
pm2.61ne.1	|- (A = B -> (ps <-> ch))
pm2.61ne.2	|- ((ph /\ A =/= B) -> ps)
pm2.61ne.3	|- (ph -> ch)

Assertion

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

Proof of Theorem pm2.61ne

Step	Hyp	Ref	Expression
1		pm2.61ne.2	. . 3 |- ((ph /\ A =/= B) -> ps)
2	1	expcom 424	. 2 |- (A =/= B -> (ph -> ps))
3		nne 2521	. . 3 |- (-. A =/= B <-> A = B)
4		pm2.61ne.3	. . . 4 |- (ph -> ch)
5		pm2.61ne.1	. . . 4 |- (A = B -> (ps <-> ch))
6	4, 5	syl5ibr 212	. . 3 |- (A = B -> (ph -> ps))
7	3, 6	sylbi 187	. 2 |- (-. A =/= B -> (ph -> ps))
8	2, 7	pm2.61i 156	1 |- (ph -> ps)

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176   /\ 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: (None)