Theorem bj-sbel1 35086

Description: Version of sbcel1g 4353 when substituting a set. (Note: one could have a corresponding version of sbcel12 4348 when substituting a set, but the point here is that the antecedent of sbcel1g 4353 is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018.)

Assertion

Ref	Expression
bj-sbel1	⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝐴(𝑥,𝑦)   𝐵(𝑦)

Proof of Theorem bj-sbel1

Step	Hyp	Ref	Expression
1		sbsbc 3724	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ [𝑦 / 𝑥]𝐴 ∈ 𝐵)
2		sbcel1g 4353	. . 3 ⊢ (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵))
3	2	elv 3437	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)
4	1, 3	bitri 274	1 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)

Syntax hints:   ↔ wb 205  [wsb 2071   ∈ wcel 2110  Vcvv 3431  [wsbc 3720  ⦋csb 3837

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2158  ax-12 2175  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1545  df-fal 1555  df-ex 1787  df-nf 1791  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-nfc 2891  df-v 3433  df-sbc 3721  df-csb 3838  df-dif 3895  df-nul 4263

This theorem is referenced by:  bj-snsetex  35149