Theorem bj-sbel1 37650

Description: Version of sbcel1g 4377 when substituting a set. (Note: one could have a corresponding version of sbcel12 4372 when substituting a set, but the point here is that the antecedent of sbcel1g 4377 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 3746	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ [𝑦 / 𝑥]𝐴 ∈ 𝐵)
2		sbcel1g 4377	. . 3 ⊢ (𝑦 ∈ V → ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵))
3	2	elv 3458	. 2 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)
4	1, 3	bitri 278	1 ⊢ ([𝑦 / 𝑥]𝐴 ∈ 𝐵 ↔ ⦋𝑦 / 𝑥⦌𝐴 ∈ 𝐵)

This proof depends on syntax axioms:   ↔ wb 209  [wsb 2099   ∈ wcel 2145  Vcvv 3453  [wsbc 3742  ⦋csb 3850

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-v 3455  df-sbc 3743  df-csb 3851  df-dif 3905  df-nul 4283

This theorem is used by:  bj-snsetex  37709