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Metamath Proof Explorer

Theorem clelsb1f

Description: Substitution for the first argument of the membership predicate in an atomic formula (class version of elsb1 ). Usage of this theorem is discouraged because it depends on ax-13 . See clelsb1fw not requiring ax-13 , but extra disjoint variables. (Contributed by Rodolfo Medina, 28-Apr-2010) (Proof shortened by Andrew Salmon, 14-Jun-2011) (Revised by Thierry Arnoux, 13-Mar-2017) (Proof shortened by Wolf Lammen, 7-May-2023) (New usage is discouraged.)

Ref Expression
Hypothesis clelsb1f.1 ⊢ Ⅎ 𝑥 𝐴
Assertion clelsb1f ( [ 𝑦 / 𝑥 ] 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 clelsb1f.1 ⊢ Ⅎ 𝑥 𝐴
2 1 nfcri ⊢ Ⅎ 𝑥 𝑤 ∈ 𝐴
3 2 sbco2 ⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑤 ] 𝑤 ∈ 𝐴 ↔ [ 𝑦 / 𝑤 ] 𝑤 ∈ 𝐴 )
4 clelsb1 ⊢ ( [ 𝑥 / 𝑤 ] 𝑤 ∈ 𝐴 ↔ 𝑥 ∈ 𝐴 )
5 4 sbbii ⊢ ( [ 𝑦 / 𝑥 ] [ 𝑥 / 𝑤 ] 𝑤 ∈ 𝐴 ↔ [ 𝑦 / 𝑥 ] 𝑥 ∈ 𝐴 )
6 clelsb1 ⊢ ( [ 𝑦 / 𝑤 ] 𝑤 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 )
7 3 5 6 3bitr3i ⊢ ( [ 𝑦 / 𝑥 ] 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 )