Theorem sbcan 3779

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 3740	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 3740	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 482	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 3733	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 3733	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 3733	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 638	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 2091	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 3505	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 379	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 207   ∧ wa 396   = wceq 1547  [wsb 2073   ∈ wcel 2119  Vcvv 3432  [wsbc 3730

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-ext 2712

This theorem depends on definitions:  df-bi 208  df-an 397  df-tru 1550  df-ex 1787  df-sb 2074  df-clab 2719  df-cleq 2732  df-clel 2815  df-v 3434  df-sbc 3731

This theorem is referenced by:  sbc3an  3794  sbcabel  3817  2nreu  4379  csbopg  4829  csbuni  4875  csbmpt12  5506  csbxp  5726  sbcfung  6516  sbcfng  6659  sbcfg  6660  fmptsnd  7120  csbfrecsg  8231  f1od2  32818  esum2dlem  34283  bnj976  34967  bnj110  35047  bnj1040  35161  csboprabg  37699  csbmpo123  37700  f1omptsnlem  37705  mptsnunlem  37707  relowlpssretop  37733  csbfinxpg  37757  sbcani  38482  sbccom2lem  38498  minregex  43985  brtrclfv2  44178  cotrclrcl  44193  frege124d  44212  sbiota1  44885  onfrALTlem5  44993  onfrALTlem4  44994  csbingVD  45334  onfrALTlem5VD  45335  onfrALTlem4VD  45336  csbxpgVD  45344  csbunigVD  45348  rspesbcd  45388