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  7193  shftfibg  11401  shftfib  11404  2shfti  11412  prdsex  13373  prdsval  13377  eqgval  13831  dvdsrd  14130  unitpropdg  14184  znleval  14689  lmbr  14964  wlkpropg  16202  wlkv  16204  wlkvg  16206  trlsfvalg  16261  trlsv  16262  eupthsg  16323  eupthv  16324