Theorem sbc3an 3752

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 1091	. . . 4 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbcbii 3743	. . 3 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sbcan 3735	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒))
4		sbcan 3735	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))
5	4	anbi1i 627	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
6	2, 3, 5	3bitri 300	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
7		df-3an 1091	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
8	6, 7	bitr4i 281	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Syntax hints:   ↔ wb 209   ∧ wa 399   ∧ w3a 1089  [wsbc 3683

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-3an 1091  df-tru 1546  df-ex 1788  df-sb 2073  df-clab 2715  df-cleq 2728  df-clel 2809  df-v 3400  df-sbc 3684

This theorem is referenced by:  bnj156  32373  bnj206  32376  bnj976  32424  bnj121  32517  bnj130  32521  bnj581  32555  bnj1040  32619  csbwrecsg  35184  topdifinffinlem  35204  rdgeqoa  35227  cdlemkid3N  38633  cdlemkid4  38634