Metamath Proof Explorer


Theorem csbfv2g

Description: Move class substitution in and out of a function value. (Contributed by NM, 10-Nov-2005)

Ref Expression
Assertion csbfv2g ( 𝐴 ∈ 𝐶 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ‘ 𝐵 ) = ( 𝐹 ‘ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 csbfv12 ⊢ ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ‘ 𝐵 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ‘ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 )
2 csbconstg ⊢ ( 𝐴 ∈ 𝐶 → ⦋ 𝐴 / 𝑥 ⦌ 𝐹 = 𝐹 )
3 2 fveq1d ⊢ ( 𝐴 ∈ 𝐶 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ‘ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ) = ( 𝐹 ‘ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ) )
4 1 3 eqtrid ⊢ ( 𝐴 ∈ 𝐶 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ‘ 𝐵 ) = ( 𝐹 ‘ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ) )