Theorem sbc3an 3461

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 1033	. . . 4 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbcbii 3458	. . 3 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sbcan 3445	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒))
4		sbcan 3445	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))
5	4	anbi1i 727	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
6	2, 3, 5	3bitri 285	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
7		df-3an 1033	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
8	6, 7	bitr4i 266	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 195   ∧ wa 383   ∧ w3a 1031  [wsbc 3402

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-v 3175  df-sbc 3403

This theorem is referenced by:  bnj156  30050  bnj206  30053  bnj976  30102  bnj121  30194  bnj130  30198  bnj581  30232  bnj1040  30294  csbwrecsg  32349  topdifinffinlem  32371  rdgeqoa  32394  cdlemkid3N  35239  cdlemkid4  35240