Metamath Proof Explorer


Theorem eqeq2d

Description: Deduction from equality to equivalence of equalities. (Contributed by NM, 27-Dec-1993) Allow shortening of eqeq2 . (Revised by Wolf Lammen, 19-Nov-2019)

Ref Expression
Hypothesis eqeq2d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion eqeq2d ( 𝜑 → ( 𝐶 = 𝐴 ↔ 𝐶 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eqeq2d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 1 eqeq1d ⊢ ( 𝜑 → ( 𝐴 = 𝐶 ↔ 𝐵 = 𝐶 ) )
3 eqcom ⊢ ( 𝐶 = 𝐴 ↔ 𝐴 = 𝐶 )
4 eqcom ⊢ ( 𝐶 = 𝐵 ↔ 𝐵 = 𝐶 )
5 2 3 4 3bitr4g ⊢ ( 𝜑 → ( 𝐶 = 𝐴 ↔ 𝐶 = 𝐵 ) )