Theorem breqd 4141

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 4132	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 105   = wceq 1402   class class class wbr 4130

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-17 1579  ax-ial 1587  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-clel 2234  df-br 4131

This theorem is used by:  breq123d  4144  breqdi  4145  sbcbr12g  4186  supeq123d  7332  shftfibg  11601  shftfib  11604  2shfti  11612  eqgval  14079  prdsex  14256  prdsval  14257  dvdsrd  14485  unitpropdg  14539  znleval  15072  lmbr  15405  wlkpropg  16731  wlkv  16733  wlkvg  16735  trlsfvalg  16790  trlsv  16791  eupthsg  16852  eupthv  16853