Theorem sylan9req 2225

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 2177	. 2 |- ( ph -> A = B )
3		sylan9req.2	. 2 |- ( ps -> B = C )
4	2, 3	sylan9eq 2224	1 |- ( ( ph /\ ps ) -> A = C )

Syntax hints:   -> wi 4   /\ wa 103   = wceq 1349

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1441  ax-gen 1443  ax-4 1504  ax-17 1520  ax-ext 2153

This theorem depends on definitions:  df-bi 116  df-cleq 2164

This theorem is referenced by:  fndmu  5301  fodmrnu  5430  funcoeqres  5476  fvunsng  5694  prarloclem5  7466  addlocprlemeq  7499  zdiv  9304  resqrexlemnm  10986  fprodssdc  11557  dvdsmulc  11785  cncongrcoprm  12064  mgmidmo  12630  lgsmodeq  13825