Theorem sbcel2 4359

Description: Move proper substitution in and out of a membership relation. (Contributed by NM, 14-Nov-2005.) (Revised by NM, 18-Aug-2018.)

Assertion

Ref	Expression
sbcel2	⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)

Distinct variable group:   𝑥,𝐵

Allowed substitution hints:   𝐴(𝑥)   𝐶(𝑥)

Proof of Theorem sbcel2

Step	Hyp	Ref	Expression
1		sbcel12 4352	. . 3 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3857	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eleq1d 2822	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrid 283	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
5		sbcex 3739	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 → 𝐴 ∈ V)
6	5	con3i 154	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ [𝐴 / 𝑥]𝐵 ∈ 𝐶)
7		noel 4279	. . . 4 ⊢ ¬ 𝐵 ∈ ∅
8		csbprc 4350	. . . . 5 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = ∅)
9	8	eleq2d 2823	. . . 4 ⊢ (¬ 𝐴 ∈ V → (𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ∅))
10	7, 9	mtbiri 327	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
11	6, 10	2falsed 376	. 2 ⊢ (¬ 𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
12	4, 11	pm2.61i 182	1 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 206   ∈ wcel 2114  Vcvv 3430  [wsbc 3729  ⦋csb 3838  ∅c0 4274

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-nfc 2886  df-v 3432  df-sbc 3730  df-csb 3839  df-dif 3893  df-nul 4275

This theorem is referenced by:  csbcom  4361  sbccsb  4377  sbnfc2  4380  csbab  4381  sbcssg  4462  csbuni  4881  csbxp  5725  csbdm  5846  issubc  17793  esum2dlem  34252  weiunlem  36661  bj-sbeq  37224  bj-sbceqgALT  37225  f1omptsnlem  37666  csbcom2fi  38463  sbcssgVD  45327  csbingVD  45328  csbunigVD  45342  disjinfi  45640  iccelpart  47905