Theorem sbcel1v 3807

Description: Class substitution into a membership relation. (Contributed by NM, 17-Aug-2018.) Avoid ax-13 2404. (Revised by Wolf Lammen, 30-Apr-2023.)

Assertion

Ref	Expression
sbcel1v	⊢ ([𝐴 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵)

Distinct variable group:   𝑥,𝐵

Allowed substitution hint:   𝐴(𝑥)

Proof of Theorem sbcel1v

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3752	. 2 ⊢ ([𝐴 / 𝑥]𝑥 ∈ 𝐵 → 𝐴 ∈ V)
2		elex 3475	. 2 ⊢ (𝐴 ∈ 𝐵 → 𝐴 ∈ V)
3		dfsbcq2 3745	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 ∈ 𝐵 ↔ [𝐴 / 𝑥]𝑥 ∈ 𝐵))
4		eleq1 2852	. . 3 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
5		clelsb1 2891	. . 3 ⊢ ([𝑦 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)
6	3, 4, 5	vtoclbg 3523	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
7	1, 2, 6	pm5.21nii 381	1 ⊢ ([𝐴 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵)

This proof depends on syntax axioms:   ↔ wb 209  [wsb 2101   ∈ wcel 2147  Vcvv 3454  [wsbc 3742

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-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-sbc 3743

This theorem is used by:  tfinds2  7902  f1od2  8138  filuni  24237  gropeld  29644  grstructeld  29645  esum2dlem  34807  bnj110  35571  f1omptsnlem  38464  relowlpssretop  38492  rdgeqoa  38498  minregex  44738  cotrclrcl  44945  frege70  45136  frege72  45138  frege91  45157  sbcoreleleq  45720  onfrALTlem4  45728  sbcoreleleqVD  46043  onfrALTlem4VD  46070  rspesbcd  46122