Theorem sbcel2 3989

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 3983	. . 3 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
2		csbconstg 3546	. . . 4 ⊢ (𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐵 = 𝐵)
3	2	eleq1d 2686	. . 3 ⊢ (𝐴 ∈ V → (⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
4	1, 3	syl5bb 272	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
5		sbcex 3445	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 → 𝐴 ∈ V)
6	5	con3i 150	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ [𝐴 / 𝑥]𝐵 ∈ 𝐶)
7		noel 3919	. . . 4 ⊢ ¬ 𝐵 ∈ ∅
8		csbprc 3980	. . . . 5 ⊢ (¬ 𝐴 ∈ V → ⦋𝐴 / 𝑥⦌𝐶 = ∅)
9	8	eleq2d 2687	. . . 4 ⊢ (¬ 𝐴 ∈ V → (𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ 𝐵 ∈ ∅))
10	7, 9	mtbiri 317	. . 3 ⊢ (¬ 𝐴 ∈ V → ¬ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)
11	6, 10	2falsed 366	. 2 ⊢ (¬ 𝐴 ∈ V → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
12	4, 11	pm2.61i 176	1 ⊢ ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ 𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶)

Syntax hints:  ¬ wn 3   ↔ wb 196   ∈ wcel 1990  Vcvv 3200  [wsbc 3435  ⦋csb 3533  ∅c0 3915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-fal 1489  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-nul 3916

This theorem is referenced by:  csbcom  3994  sbccsb  4004  sbnfc2  4007  csbab  4008  sbcssg  4085  csbuni  4466  csbxp  5200  csbdm  5318  issubc  16495  esum2dlem  30154  bj-sbeq  32896  bj-sbceqgALT  32897  bj-sels  32950  f1omptsnlem  33183  csbcom2fi  33934  disjinfi  39380  iccelpart  41369