Metamath Proof Explorer


Theorem fveqeq2

Description: Equality deduction for function value. (Contributed by BJ, 31-Aug-2022)

Ref Expression
Assertion fveqeq2 ( 𝐴 = 𝐵 → ( ( 𝐹𝐴 ) = 𝐶 ↔ ( 𝐹𝐵 ) = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 id ( 𝐴 = 𝐵𝐴 = 𝐵 )
2 1 fveqeq2d ( 𝐴 = 𝐵 → ( ( 𝐹𝐴 ) = 𝐶 ↔ ( 𝐹𝐵 ) = 𝐶 ) )