Theorem bj-sbel1 37338

Description: Version of sbcel1g 4364 when substituting a set. (Note: one could have a corresponding version of sbcel12 4359 when substituting a set, but the point here is that the antecedent of sbcel1g 4364 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 3743	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ [𝑦 / 𝑥]𝐴 ∈ 𝐵)
2		sbcel1g 4364	. . 3 ⊢ (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵))
3	2	elv 3453	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)
4	1, 3	bitri 277	1 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)

Syntax hints:   ↔ wb 208  [wsb 2084   ∈ wcel 2136  Vcvv 3448  [wsbc 3739  ⦋csb 3847

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-8 2138  ax-9 2146  ax-10 2169  ax-11 2185  ax-12 2206  ax-ext 2728

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 857  df-tru 1557  df-fal 1567  df-ex 1794  df-nf 1798  df-sb 2085  df-clab 2735  df-cleq 2748  df-clel 2831  df-nfc 2905  df-v 3450  df-sbc 3740  df-csb 3848  df-dif 3902  df-nul 4281

This theorem is referenced by:  bj-snsetex  37396