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  7331  shftfibg  11585  shftfib  11588  2shfti  11596  eqgval  14026  prdsex  14172  prdsval  14173  dvdsrd  14401  unitpropdg  14455  znleval  14988  lmbr  15314  wlkpropg  16565  wlkv  16567  wlkvg  16569  trlsfvalg  16624  trlsv  16625  eupthsg  16686  eupthv  16687