Metamath Proof Explorer


Theorem fvifeq

Description: Equality of function values with conditional arguments, see also fvif . (Contributed by Alexander van der Vekens, 21-May-2018)

Ref Expression
Assertion fvifeq ( 𝐴 = if ( 𝜑 , 𝐵 , 𝐶 ) → ( 𝐹 ‘ 𝐴 ) = if ( 𝜑 , ( 𝐹 ‘ 𝐵 ) , ( 𝐹 ‘ 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ ( 𝐴 = if ( 𝜑 , 𝐵 , 𝐶 ) → ( 𝐹 ‘ 𝐴 ) = ( 𝐹 ‘ if ( 𝜑 , 𝐵 , 𝐶 ) ) )
2 fvif ⊢ ( 𝐹 ‘ if ( 𝜑 , 𝐵 , 𝐶 ) ) = if ( 𝜑 , ( 𝐹 ‘ 𝐵 ) , ( 𝐹 ‘ 𝐶 ) )
3 1 2 eqtrdi ⊢ ( 𝐴 = if ( 𝜑 , 𝐵 , 𝐶 ) → ( 𝐹 ‘ 𝐴 ) = if ( 𝜑 , ( 𝐹 ‘ 𝐵 ) , ( 𝐹 ‘ 𝐶 ) ) )