Metamath Proof Explorer


Theorem fveq1

Description: Equality theorem for function value. (Contributed by NM, 29-Dec-1996)

Ref Expression
Assertion fveq1 ( 𝐹 = 𝐺 → ( 𝐹 ‘ 𝐴 ) = ( 𝐺 ‘ 𝐴 ) )

Proof

Step Hyp Ref Expression
1 breq ⊢ ( 𝐹 = 𝐺 → ( 𝐴 𝐹 𝑥 ↔ 𝐴 𝐺 𝑥 ) )
2 1 iotabidv ⊢ ( 𝐹 = 𝐺 → ( ℩ 𝑥 𝐴 𝐹 𝑥 ) = ( ℩ 𝑥 𝐴 𝐺 𝑥 ) )
3 df-fv ⊢ ( 𝐹 ‘ 𝐴 ) = ( ℩ 𝑥 𝐴 𝐹 𝑥 )
4 df-fv ⊢ ( 𝐺 ‘ 𝐴 ) = ( ℩ 𝑥 𝐴 𝐺 𝑥 )
5 2 3 4 3eqtr4g ⊢ ( 𝐹 = 𝐺 → ( 𝐹 ‘ 𝐴 ) = ( 𝐺 ‘ 𝐴 ) )