Theorem sbcan 3821

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 3782	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 3782	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 484	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 3775	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 3775	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 3775	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 632	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 2082	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 3569	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 382	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1533  [wsb 2065   ∈ wcel 2110  Vcvv 3495  [wsbc 3772

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-12 2172  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-v 3497  df-sbc 3773

This theorem is referenced by:  sbc3an  3838  sbcabel  3861  2nreu  4393  csbopg  4815  csbuni  4860  csbmpt12  5437  csbxp  5645  difopab  5697  sbcfung  6374  sbcfng  6506  sbcfg  6507  fmptsnd  6926  f1od2  30451  esum2dlem  31346  bnj976  32044  bnj110  32125  bnj1040  32239  csbwrecsg  34602  csboprabg  34605  csbmpo123  34606  f1omptsnlem  34611  mptsnunlem  34613  relowlpssretop  34639  csbfinxpg  34663  sbcani  35380  sbccom2lem  35396  brtrclfv2  40065  cotrclrcl  40080  frege124d  40099  sbiota1  40759  onfrALTlem5  40869  onfrALTlem4  40870  csbingVD  41211  onfrALTlem5VD  41212  onfrALTlem4VD  41213  csbxpgVD  41221  csbunigVD  41225