Metamath Proof Explorer


Theorem fveq1d

Description: Equality deduction for function value. (Contributed by NM, 2-Sep-2003)

Ref Expression
Hypothesis fveq1d.1 ⊢ φ → F = G
Assertion fveq1d ⊢ φ → F ⁡ A = G ⁡ A

Proof

Step Hyp Ref Expression
1 fveq1d.1 ⊢ φ → F = G
2 fveq1 ⊢ F = G → F ⁡ A = G ⁡ A
3 1 2 syl ⊢ φ → F ⁡ A = G ⁡ A