Theorem breqd 4097

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 4088	. 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 1395   class class class wbr 4086

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 1493  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-4 1556  ax-17 1572  ax-ial 1580  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-cleq 2222  df-clel 2225  df-br 4087

This theorem is referenced by:  breq123d  4100  breqdi  4101  sbcbr12g  4142  supeq123d  7181  shftfibg  11371  shftfib  11374  2shfti  11382  prdsex  13342  prdsval  13346  eqgval  13800  dvdsrd  14098  unitpropdg  14152  znleval  14657  lmbr  14927  wlkpropg  16121  wlkv  16123  wlkvg  16125  trlsfvalg  16178  trlsv  16179  eupthsg  16240  eupthv  16241