Theorem 3netr3g 2545

Description: Substitution of equality into both sides of an inequality. (Contributed by NM, 24-Jul-2012.)

Hypotheses

Ref	Expression
3netr3g.1	|- (ph -> A =/= B)
3netr3g.2	|- A = C
3netr3g.3	|- B = D

Assertion

Ref	Expression
3netr3g	|- (ph -> C =/= D)

Proof of Theorem 3netr3g

Step	Hyp	Ref	Expression
1		3netr3g.1	. 2 |- (ph -> A =/= B)
2		3netr3g.2	. . 3 |- A = C
3		3netr3g.3	. . 3 |- B = D
4	2, 3	neeq12i 2529	. 2 |- (A =/= B <-> C =/= D)
5	1, 4	sylib 188	1 |- (ph -> C =/= D)

Syntax hints:   -> wi 4   = wceq 1642   =/= wne 2517

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 2519

This theorem is referenced by: (None)