Theorem pm2.61dne 2594

Description: Deduction eliminating an inequality in an antecedent. (Contributed by NM, 1-Jun-2007.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Hypotheses

Ref	Expression
pm2.61dne.1	|- (ph -> (A = B -> ps))
pm2.61dne.2	|- (ph -> (A =/= B -> ps))

Assertion

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

Proof of Theorem pm2.61dne

Step	Hyp	Ref	Expression
1		pm2.61dne.2	. 2 |- (ph -> (A =/= B -> ps))
2		nne 2521	. . 3 |- (-. A =/= B <-> A = B)
3		pm2.61dne.1	. . 3 |- (ph -> (A = B -> ps))
4	2, 3	syl5bi 208	. 2 |- (ph -> (-. A =/= B -> ps))
5	1, 4	pm2.61d 150	1 |- (ph -> ps)

This proof depends on syntax axioms:  -. wn 3   -> wi 4   = wceq 1642   =/= wne 2517

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-ne 2519

This theorem is used by:  pm2.61dane  2595