Theorem sbcan 3802

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016.) (Revised by NM, 17-Aug-2018.)

Assertion

Ref	Expression
sbcan	⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Proof of Theorem sbcan

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3763	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 3763	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 486	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 3756	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 3756	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 3756	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 643	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 2120	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 3533	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 381	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 209   ∧ wa 400   = wceq 1567  [wsb 2097   ∈ wcel 2149  Vcvv 3463  [wsbc 3753

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-v 3465  df-sbc 3754

This theorem is referenced by:  sbc3an  3817  sbcabel  3840  2nreu  4407  csbopg  4857  csbuni  4904  csbmpt12  5540  csbxp  5760  sbcfung  6557  sbcfng  6700  sbcfg  6701  fmptsnd  7165  csbfrecsg  8277  f1od2  33001  esum2dlem  34423  bnj976  35107  bnj110  35187  bnj1040  35301  csboprabg  37859  csbmpo123  37860  f1omptsnlem  37865  mptsnunlem  37867  relowlpssretop  37893  csbfinxpg  37917  sbcani  38642  sbccom2lem  38658  minregex  44145  brtrclfv2  44338  cotrclrcl  44353  frege124d  44372  sbiota1  45029  onfrALTlem5  45136  onfrALTlem4  45137  csbingVD  45477  onfrALTlem5VD  45478  onfrALTlem4VD  45479  csbxpgVD  45487  csbunigVD  45491  rspesbcd  45531