Theorem breqd 4009

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 4000	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1353   class class class wbr 3998

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 1445  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-4 1508  ax-17 1524  ax-ial 1532  ax-ext 2157

This theorem depends on definitions:  df-bi 117  df-cleq 2168  df-clel 2171  df-br 3999

This theorem is referenced by:  breq123d  4012  breqdi  4013  sbcbr12g  4053  supeq123d  6980  shftfibg  10795  shftfib  10798  2shfti  10806  lmbr  13264