Theorem breqd 4009

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 4000	. 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 1353   class class class wbr 3998

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 1445  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-4 1508  ax-17 1524  ax-ial 1532  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-cleq 2168  df-clel 2171  df-br 3999

This theorem is referenced by:  breq123d  4012  breqdi  4013  sbcbr12g  4053  supeq123d  6980  shftfibg  10797  shftfib  10800  2shfti  10808  lmbr  13284