Metamath Proof Explorer


Theorem csbov1g

Description: Move class substitution in and out of an operation. (Contributed by NM, 12-Nov-2005)

Ref Expression
Assertion csbov1g ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 𝐹 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝐹 𝐶 ) )

Proof

Step Hyp Ref Expression
1 csbov12g ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 𝐹 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝐹 ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )
2 csbconstg ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = 𝐶 )
3 2 oveq2d ⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝐹 ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝐹 𝐶 ) )
4 1 3 eqtrd ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐵 𝐹 𝐶 ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 𝐹 𝐶 ) )