Theorem eqeqan2d 38949

Description: Implication of introducing a new equality. (Contributed by Peter Mazsa, 17-Apr-2019.)

Hypothesis

Ref	Expression
eqeqan2d.1	⊢ (𝜑 → 𝐶 = 𝐷)

Assertion

Ref	Expression
eqeqan2d	⊢ ((𝐴 = 𝐵 ∧ 𝜑) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Proof of Theorem eqeqan2d

Step	Hyp	Ref	Expression
1		eqeqan2d.1	. 2 ⊢ (𝜑 → 𝐶 = 𝐷)
2		eqeq12 2782	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
3	1, 2	sylan2 605	1 ⊢ ((𝐴 = 𝐵 ∧ 𝜑) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2757

This theorem is used by:  ecqmap  39156