Metamath Proof Explorer


Theorem sbcel1v

Description: Class substitution into a membership relation. (Contributed by NM, 17-Aug-2018) Avoid ax-13 . (Revised by Wolf Lammen, 30-Apr-2023)

Ref Expression
Assertion sbcel1v ( [ 𝐴 / 𝑥 ] 𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵 )

Proof

Step Hyp Ref Expression
1 sbcex ⊢ ( [ 𝐴 / 𝑥 ] 𝑥 ∈ 𝐵 → 𝐴 ∈ V )
2 elex ⊢ ( 𝐴 ∈ 𝐵 → 𝐴 ∈ V )
3 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝑥 ∈ 𝐵 ↔ [ 𝐴 / 𝑥 ] 𝑥 ∈ 𝐵 ) )
4 eleq1 ⊢ ( 𝑦 = 𝐴 → ( 𝑦 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵 ) )
5 clelsb1 ⊢ ( [ 𝑦 / 𝑥 ] 𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵 )
6 3 4 5 vtoclbg ⊢ ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] 𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵 ) )
7 1 2 6 pm5.21nii ⊢ ( [ 𝐴 / 𝑥 ] 𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵 )