Theorem sbcel2 4379

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 4372	. . 3 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3870	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eleq1d 2849	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	bitrid 286	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
5		sbcex 3752	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 → 𝐴 ∈ V)
6	5	con3i 155	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ [𝐴 / 𝑥]𝐵 ∈ 𝐶)
7		noel 4286	. . . 4 ⊢ ¬ 𝐵 ∈ ∅
8		csbprc 4370	. . . . 5 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = ∅)
9	8	eleq2d 2850	. . . 4 ⊢ (¬ 𝐴 ∈ V → (𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ∅))
10	7, 9	mtbiri 330	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
11	6, 10	2falsed 379	. 2 ⊢ (¬ 𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
12	4, 11	pm2.61i 184	1 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∈ wcel 2147  Vcvv 3454  [wsbc 3742  ⦋csb 3851  ∅c0 4281

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-v 3456  df-sbc 3743  df-csb 3852  df-dif 4050  df-nul 4282

This theorem is used by:  csbcom  4381  sbccsb  4397  sbnfc2  4400  csbab  4401  sbcssg  4480  csbuni  4901  csbxp  5758  csbdm  5887  issubc  18042  esum2dlem  34807  weiunlem  37442  bj-sbeq  38004  bj-sbceqgALT  38005  f1omptsnlem  38464  csbcom2fi  39263  sbcssgVD  46067  csbingVD  46068  csbunigVD  46082  disjinfi  46422  iccelpart  48730