Theorem eqeqan2d

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

	Ref	Expression
Hypothesis	eqeqan2d.1	⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion	eqeqan2d	⊢ ( ( 𝐴 = 𝐵 ∧ 𝜑 ) → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐷 ) )

Proof

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