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Metamath Proof Explorer

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 ) )