Theorem eqeqan1d 34003

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

Hypothesis

Ref	Expression
eqeqan1d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
eqeqan1d	⊢ ((𝜑 ∧ 𝐶 = 𝐷) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Proof of Theorem eqeqan1d

Step	Hyp	Ref	Expression
1		eqeqan1d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		eqeq12 2635	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))
3	1, 2	sylan 488	1 ⊢ ((𝜑 ∧ 𝐶 = 𝐷) → (𝐴 = 𝐶 ↔ 𝐵 = 𝐷))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384   = wceq 1483

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-an 386  df-ex 1705  df-cleq 2615

This theorem is referenced by: (None)