Theorem sylan9req 2090

Description: An equality transitivity deduction. (Contributed by NM, 23-Jun-2007.)

Hypotheses

Ref	Expression
sylan9req.1	|- (ph -> B = A)
sylan9req.2	|- (ps -> B = C)

Assertion

Ref	Expression
sylan9req	|- ((ph /\ ps) -> A = C)

Proof of Theorem sylan9req

Step	Hyp	Ref	Expression
1		sylan9req.1	. . 3 |- (ph -> B = A)
2	1	eqcomd 2042	. 2 |- (ph -> A = B)
3		sylan9req.2	. 2 |- (ps -> B = C)
4	2, 3	sylan9eq 2089	1 |- ((ph /\ ps) -> A = C)

Syntax hints:   -> wi 4   /\ wa 97   = wceq 1242

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-4 1397  ax-17 1416  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-cleq 2030

This theorem is referenced by:  xpid11m  4500  fndmu  4943  fodmrnu  5057  funcoeqres  5100  fvunsng  5300  prarloclem5  6483  addlocprlemeq  6516  zdiv  8104