Theorem sbc3an 3848

Description: Distribution of class substitution over triple conjunction. (Contributed by NM, 14-Dec-2006.) (Revised by NM, 17-Aug-2018.)

Assertion

Ref	Expression
sbc3an	⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3an

Step	Hyp	Ref	Expression
1		df-3an 1090	. . . 4 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbcbii 3838	. . 3 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sbcan 3830	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒))
4		sbcan 3830	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))
5	4	anbi1i 625	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
6	2, 3, 5	3bitri 297	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
7		df-3an 1090	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
8	6, 7	bitr4i 278	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 205   ∧ wa 397   ∧ w3a 1088  [wsbc 3778

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-3an 1090  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-sbc 3779

This theorem is referenced by:  csbfrecsg  8269  bnj156  33770  bnj206  33773  bnj976  33819  bnj121  33912  bnj130  33916  bnj581  33950  bnj1040  34014  topdifinffinlem  36276  rdgeqoa  36299  cdlemkid3N  39852  cdlemkid4  39853  minregex  42333