Theorem breqd 4125

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

Hypothesis

Ref	Expression
breq1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
breqd	⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Proof of Theorem breqd

Step	Hyp	Ref	Expression
1		breq1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		breq 4116	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1398   class class class wbr 4114

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 1496  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-4 1559  ax-17 1575  ax-ial 1583  ax-ext 2216

This theorem depends on definitions:  df-bi 117  df-cleq 2227  df-clel 2230  df-br 4115

This theorem is referenced by:  breq123d  4128  breqdi  4129  sbcbr12g  4170  supeq123d  7295  shftfibg  11530  shftfib  11533  2shfti  11541  eqgval  13976  prdsex  14114  prdsval  14115  dvdsrd  14339  unitpropdg  14393  znleval  14927  lmbr  15204  wlkpropg  16445  wlkv  16447  wlkvg  16449  trlsfvalg  16504  trlsv  16505  eupthsg  16566  eupthv  16567