Metamath Proof Explorer


Theorem sbcel1g

Description: Move proper substitution in and out of a membership relation. Note that the scope of [. A / x ]. is the wff B e. C , whereas the scope of [_ A / x ]_ is the class B . (Contributed by NM, 10-Nov-2005)

Ref Expression
Assertion sbcel1g ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 sbcel12 ⊢ ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 )
2 csbconstg ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = 𝐶 )
3 2 eleq2d ⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )
4 1 3 bitrid ⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )