Theorem 3anasss 1380

Description: Associative law for conjunction applied to antecedent (eliminates syllogism). Converse of 3anassrs 1381. (Contributed by Thierry Arnoux, 5-Jul-2026.)

Hypothesis

Ref	Expression
3anasss.1	⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃) → 𝜏)

Assertion

Ref	Expression
3anasss	⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)

Proof of Theorem 3anasss

Step	Hyp	Ref	Expression
1		13an22anass 1379	. 2 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) ↔ ((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃)))
2		3anasss.1	. . 3 ⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) ∧ 𝜃) → 𝜏)
3	2	anasss 472	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ (𝜒 ∧ 𝜃)) → 𝜏)
4	1, 3	sylbi 220	1 ⊢ ((𝜑 ∧ (𝜓 ∧ 𝜒 ∧ 𝜃)) → 𝜏)

This proof depends on syntax axioms:   → wi 4   ∧ 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:  cgrarag  29337  ragsupplcgra  29338  angmgmaddeu1  29372  angmgmaddeu2  29373  angmgmaddeu3  29374  angmgmaddov2lem  29380  angmgmaddcpbl  29383  prlngmolem1  29423