Metamath Proof Explorer

Theorem feq2d

Description: Equality deduction for functions. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis feq2d.1 ⊢ φ → A = B
Assertion feq2d ⊢ φ → F : A ⟶ C ↔ F : B ⟶ C

Proof

Step Hyp Ref Expression
1 feq2d.1 ⊢ φ → A = B
2 feq2 ⊢ A = B → F : A ⟶ C ↔ F : B ⟶ C
3 1 2 syl ⊢ φ → F : A ⟶ C ↔ F : B ⟶ C