Theorem neeq1d 2529

Description: Deduction for inequality. (Contributed by NM, 25-Oct-1999.)

Hypothesis

Ref	Expression
neeq1d.1	|- (ph -> A = B)

Assertion

Ref	Expression
neeq1d	|- (ph -> (A =/= C <-> B =/= C))

Proof of Theorem neeq1d

Step	Hyp	Ref	Expression
1		neeq1d.1	. 2 |- (ph -> A = B)
2		neeq1 2524	. 2 |- (A = B -> (A =/= C <-> B =/= C))
3	1, 2	syl 15	1 |- (ph -> (A =/= C <-> B =/= C))

Syntax hints:   -> wi 4   <-> wb 176   = wceq 1642   =/= wne 2516

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-ex 1542  df-cleq 2346  df-ne 2518

This theorem is referenced by:  neeq12d  2531  eqnetrd  2534  prnzg  3836  preaddccan2lem1  4454  preaddccan2  4455  evenodddisj  4516  vfinncvntnn  4548  ereldm  5971  map0  6025  ce0addcnnul  6179  ce0nn  6180  ce0nnulb  6182