Theorem bj-sbel1 35069

Description: Version of sbcel1g 4352 when substituting a set. (Note: one could have a corresponding version of sbcel12 4347 when substituting a set, but the point here is that the antecedent of sbcel1g 4352 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 3723	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ [𝑦 / 𝑥]𝐴 ∈ 𝐵)
2		sbcel1g 4352	. . 3 ⊢ (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵))
3	2	elv 3436	. 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 2070   ∈ wcel 2109  Vcvv 3430  [wsbc 3719  ⦋csb 3836

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-10 2140  ax-11 2157  ax-12 2174  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-fal 1554  df-ex 1786  df-nf 1790  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-nfc 2890  df-v 3432  df-sbc 3720  df-csb 3837  df-dif 3894  df-nul 4262

This theorem is referenced by:  bj-snsetex  35132