Theorem 3jaob 1427

Description: Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011.) (Proof shortened by Hongxiu Chen, 29-Jun-2025.)

Assertion

Ref	Expression
3jaob	⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))

Proof of Theorem 3jaob

Step	Hyp	Ref	Expression
1		pm5.53 1006	. 2 ⊢ ((((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓) ↔ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓)) ∧ (𝜃 → 𝜓)))
2		df-3or 1087	. . 3 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜃))
3	2	imbi1i 349	. 2 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓))
4		df-3an 1088	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) ↔ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓)) ∧ (𝜃 → 𝜓)))
5	1, 3, 4	3bitr4i 303	1 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   ∨ wo 847   ∨ w3o 1085   ∧ w3a 1086

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3or 1087  df-3an 1088

This theorem is referenced by: (None)