Theorem sbc3an 3016

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 975	. . . 4 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbcbii 3014	. . 3 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sbcan 2997	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒))
4		sbcan 2997	. . . 4 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))
5	4	anbi1i 455	. . 3 ⊢ (([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
6	2, 3, 5	3bitri 205	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
7		df-3an 975	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
8	6, 7	bitr4i 186	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Syntax hints:   ∧ wa 103   ↔ wb 104   ∧ w3a 973  [wsbc 2955

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-sbc 2956

This theorem is referenced by: (None)