Theorem breqd 3935

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 3926	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1331   class class class wbr 3924

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-4 1487  ax-17 1506  ax-ial 1514  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-cleq 2130  df-clel 2133  df-br 3925

This theorem is referenced by:  breq123d  3938  breqdi  3939  sbcbr12g  3978  supeq123d  6871  shftfibg  10585  shftfib  10588  2shfti  10596  lmbr  12371