Theorem breqd 4122

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 4113	. 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 4111

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 4112

This theorem is referenced by:  breq123d  4125  breqdi  4126  sbcbr12g  4167  supeq123d  7284  shftfibg  11509  shftfib  11512  2shfti  11520  prdsex  13499  prdsval  13503  eqgval  13957  dvdsrd  14256  unitpropdg  14310  znleval  14818  lmbr  15095  wlkpropg  16336  wlkv  16338  wlkvg  16340  trlsfvalg  16395  trlsv  16396  eupthsg  16457  eupthv  16458