Theorem breqd 4113

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 4104	. 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 4102

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 2214

This theorem depends on definitions:  df-bi 117  df-cleq 2225  df-clel 2228  df-br 4103

This theorem is referenced by:  breq123d  4116  breqdi  4117  sbcbr12g  4158  supeq123d  7273  shftfibg  11483  shftfib  11486  2shfti  11494  prdsex  13456  prdsval  13460  eqgval  13914  dvdsrd  14213  unitpropdg  14267  znleval  14773  lmbr  15048  wlkpropg  16289  wlkv  16291  wlkvg  16293  trlsfvalg  16348  trlsv  16349  eupthsg  16410  eupthv  16411