Theorem sbcel2 4376

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 4369	. . 3 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3866	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eleq1d 2846	. . 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 3749	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 → 𝐴 ∈ V)
6	5	con3i 155	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ [𝐴 / 𝑥]𝐵 ∈ 𝐶)
7		noel 4284	. . . 4 ⊢ ¬ 𝐵 ∈ ∅
8		csbprc 4367	. . . . 5 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = ∅)
9	8	eleq2d 2847	. . . 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 2145  Vcvv 3451  [wsbc 3739  ⦋csb 3847  ∅c0 4279

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 2213  ax-ext 2733

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 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-nul 4280

This theorem is used by:  csbcom  4378  sbccsb  4394  sbnfc2  4397  csbab  4398  sbcssg  4477  csbuni  4898  csbxp  5752  csbdm  5879  issubc  17990  esum2dlem  34706  weiunlem  37221  bj-sbeq  37783  bj-sbceqgALT  37784  f1omptsnlem  38227  csbcom2fi  39028  sbcssgVD  45824  csbingVD  45825  csbunigVD  45839  disjinfi  46150  iccelpart  48459