Theorem bj-sbel1 36848

Description: Version of sbcel1g 4421 when substituting a set. (Note: one could have a corresponding version of sbcel12 4416 when substituting a set, but the point here is that the antecedent of sbcel1g 4421 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 3795	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ [𝑦 / 𝑥]𝐴 ∈ 𝐵)
2		sbcel1g 4421	. . 3 ⊢ (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵))
3	2	elv 3482	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)
4	1, 3	bitri 275	1 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)

Syntax hints:   ↔ wb 206  [wsb 2060   ∈ wcel 2104  Vcvv 3477  [wsbc 3791  ⦋csb 3908

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1963  ax-7 2003  ax-8 2106  ax-9 2114  ax-10 2137  ax-11 2153  ax-12 2173  ax-ext 2704

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-tru 1538  df-fal 1548  df-ex 1775  df-nf 1779  df-sb 2061  df-clab 2711  df-cleq 2725  df-clel 2812  df-nfc 2888  df-v 3479  df-sbc 3792  df-csb 3909  df-dif 3966  df-nul 4340

This theorem is referenced by:  bj-snsetex  36906