Theorem neneqd 2533

Description: Deduction eliminating inequality definition. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)

Hypothesis

Ref	Expression
neneqd.1	|- (ph -> A =/= B)

Assertion

Ref	Expression
neneqd	|- (ph -> -. A = B)

Proof of Theorem neneqd

Step	Hyp	Ref	Expression
1		neneqd.1	. 2 |- (ph -> A =/= B)
2		df-ne 2519	. 2 |- (A =/= B <-> -. A = B)
3	1, 2	sylib 188	1 |- (ph -> -. A = B)

Syntax hints:  -. wn 3   -> wi 4   = 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-ne 2519

This theorem is referenced by:  necon2bi  2563  necon2i  2564  pm2.21ddne  2591  nulnnn  4557  enprmaplem3  6079