Theorem sbcan 3750

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 3716	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 3716	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 482	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 3709	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 3709	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 3709	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 630	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 2059	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 3511	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 380	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 207   ∧ wa 396   = wceq 1522  [wsb 2042   ∈ wcel 2081  Vcvv 3437  [wsbc 3706

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1777  ax-4 1791  ax-5 1888  ax-6 1947  ax-7 1992  ax-8 2083  ax-9 2091  ax-12 2141  ax-ext 2769

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1762  df-nf 1766  df-sb 2043  df-clab 2776  df-cleq 2788  df-clel 2863  df-v 3439  df-sbc 3707

This theorem is referenced by:  sbc3an  3766  sbcabel  3789  2nreu  4307  csbopg  4728  csbuni  4773  csbmpt12  5332  csbxp  5536  difopab  5588  sbcfung  6249  sbcfng  6379  sbcfg  6380  fmptsnd  6794  f1od2  30145  esum2dlem  30968  bnj976  31666  bnj110  31746  bnj1040  31858  csbwrecsg  34139  csboprabg  34142  csbmpo123  34143  f1omptsnlem  34148  mptsnunlem  34150  relowlpssretop  34176  csbfinxpg  34200  sbcani  34918  sbccom2lem  34934  brtrclfv2  39557  cotrclrcl  39572  frege124d  39591  sbiota1  40304  onfrALTlem5  40415  onfrALTlem4  40416  csbingVD  40757  onfrALTlem5VD  40758  onfrALTlem4VD  40759  csbxpgVD  40767  csbunigVD  40771