Theorem breqd 4139

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 4130	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1402   class class class wbr 4128

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 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-4 1563  ax-17 1579  ax-ial 1587  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-cleq 2231  df-clel 2234  df-br 4129

This theorem is referenced by:  breq123d  4142  breqdi  4143  sbcbr12g  4184  supeq123d  7324  shftfibg  11566  shftfib  11569  2shfti  11577  eqgval  14006  prdsex  14152  prdsval  14153  dvdsrd  14377  unitpropdg  14431  znleval  14963  lmbr  15240  wlkpropg  16482  wlkv  16484  wlkvg  16486  trlsfvalg  16541  trlsv  16542  eupthsg  16603  eupthv  16604