Theorem sbcan 3628

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 3595	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 3595	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 467	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 3588	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 3588	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 3588	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 608	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 2545	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 3416	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 367	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 196   ∧ wa 382   = wceq 1630  [wsb 2048   ∈ wcel 2144  Vcvv 3349  [wsbc 3585

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-9 2153  ax-10 2173  ax-12 2202  ax-13 2407  ax-ext 2750

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-clab 2757  df-cleq 2763  df-clel 2766  df-v 3351  df-sbc 3586

This theorem is referenced by:  sbc3an  3643  sbcabel  3664  csbopg  4555  csbuni  4600  csbmpt12  5143  csbxp  5340  difopab  5392  sbcfung  6055  sbcfng  6182  sbcfg  6183  fmptsnd  6578  f1od2  29833  esum2dlem  30488  bnj976  31180  bnj110  31260  bnj1040  31372  csbwrecsg  33503  csboprabg  33506  csbmpt22g  33507  f1omptsnlem  33513  mptsnunlem  33515  relowlpssretop  33542  csbfinxpg  33555  sbcani  34235  sbccom2lem  34254  brtrclfv2  38538  cotrclrcl  38553  frege124d  38572  sbiota1  39154  onfrALTlem5  39276  onfrALTlem4  39277  csbxpgOLD  39570  onfrALTlem5VD  39637