Theorem breqd 4055

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 4046	. 2 ⊢ (𝐴 = 𝐵 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐶𝐴𝐷 ↔ 𝐶𝐵𝐷))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1373   class class class wbr 4044

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 1470  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-4 1533  ax-17 1549  ax-ial 1557  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-cleq 2198  df-clel 2201  df-br 4045

This theorem is referenced by:  breq123d  4058  breqdi  4059  sbcbr12g  4099  supeq123d  7093  shftfibg  11131  shftfib  11134  2shfti  11142  prdsex  13101  prdsval  13105  eqgval  13559  dvdsrd  13856  unitpropdg  13910  znleval  14415  lmbr  14685