Theorem 3an4anass 1122

Description: Associative law for four conjunctions with a triple conjunction. (Contributed by Alexander van der Vekens, 24-Jun-2018.)

Assertion

Ref	Expression
3an4anass	⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) ↔ ((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃)))

Proof of Theorem 3an4anass

Step	Hyp	Ref	Expression
1		df-3an 1105	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	anbi1i 636	. 2 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) ↔ (((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃))
3		anass 474	. 2 ⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃) ↔ ((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃)))
4	2, 3	bitri 278	1 ⊢ (((𝜑 ∧ 𝜓 ∧ 𝜒) ∧ 𝜃) ↔ ((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃)))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∧ w3a 1103

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105

This theorem is used by:  13an22anass  1379  4anpull2  1382  eqvinot  5456  cotsexgw  5462  mpt3eqdv  7686  oeeui  8611  isclwlkupgr  30365  clwlkclwwlk  30593  bnj557  35531  cantnf2  44327  3an4ancom24  48339  isthincd2  50545