Theorem breqd 4000

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 3991	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1348   class class class wbr 3989

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 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-4 1503  ax-17 1519  ax-ial 1527  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-cleq 2163  df-clel 2166  df-br 3990

This theorem is referenced by:  breq123d  4003  breqdi  4004  sbcbr12g  4044  supeq123d  6968  shftfibg  10784  shftfib  10787  2shfti  10795  lmbr  13007