Metamath Proof Explorer


Theorem sbcel2gv

Description: Class substitution into a membership relation. (Contributed by NM, 17-Nov-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Assertion sbcel2gv ( 𝐵 ∈ 𝑉 → ( [ 𝐵 / 𝑥 ] 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 eleq2 ⊢ ( 𝑥 = 𝑦 → ( 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝑦 ) )
2 eleq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 ∈ 𝑦 ↔ 𝐴 ∈ 𝐵 ) )
3 1 2 sbcie2g ⊢ ( 𝐵 ∈ 𝑉 → ( [ 𝐵 / 𝑥 ] 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵 ) )