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