Theorem breqd 4142

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 4133	. 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 4131

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 4132

This theorem is used by:  breq123d  4145  breqdi  4146  sbcbr12g  4187  supeq123d  7337  shftfibg  11606  shftfib  11609  2shfti  11617  eqgval  14084  prdsex  14278  prdsval  14279  dvdsrd  14509  unitpropdg  14563  znleval  15096  lmbr  15456  wlkpropg  16787  wlkv  16789  wlkvg  16791  trlsfvalg  16846  trlsv  16847  eupthsg  16908  eupthv  16909