Theorem breqd 4099

Description: Equality deduction for a binary relation. (Contributed by NM, 29-Oct-2011.)

Hypothesis

Ref	Expression
breq1d.1	|- ( ph -> A = B )

Assertion

Ref	Expression
breqd	|- ( ph -> ( C A D <-> C B D ) )

Proof of Theorem breqd

Step	Hyp	Ref	Expression
1		breq1d.1	. 2 |- ( ph -> A = B )
2		breq 4090	. 2 |- ( A = B -> ( C A D <-> C B D ) )
3	1, 2	syl 14	1 |- ( ph -> ( C A D <-> C B D ) )

Syntax hints:   -> wi 4   <-> wb 105   = wceq 1397   class class class wbr 4088

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 1495  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-4 1558  ax-17 1574  ax-ial 1582  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-cleq 2224  df-clel 2227  df-br 4089

This theorem is referenced by:  breq123d  4102  breqdi  4103  sbcbr12g  4144  supeq123d  7190  shftfibg  11381  shftfib  11384  2shfti  11392  prdsex  13353  prdsval  13357  eqgval  13811  dvdsrd  14110  unitpropdg  14164  znleval  14669  lmbr  14939  wlkpropg  16177  wlkv  16179  wlkvg  16181  trlsfvalg  16236  trlsv  16237  eupthsg  16298  eupthv  16299