Theorem sylan9req 2250

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 2202	. 2 |- ( ph -> A = B )
3		sylan9req.2	. 2 |- ( ps -> B = C )
4	2, 3	sylan9eq 2249	1 |- ( ( ph /\ ps ) -> A = C )

Syntax hints:   -> wi 4   /\ wa 104   = wceq 1364

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463  ax-4 1524  ax-17 1540  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-cleq 2189

This theorem is referenced by:  fndmu  5362  fodmrnu  5491  funcoeqres  5538  fvunsng  5759  prarloclem5  7584  addlocprlemeq  7617  zdiv  9431  resqrexlemnm  11200  fprodssdc  11772  dvdsmulc  12001  cncongrcoprm  12299  mgmidmo  13074  lgsmodeq  15370