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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