Theorem 3jaob 1453

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 1022	. 2 ⊢ ((((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓) ↔ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓)) ∧ (𝜃 → 𝜓)))
2		df-3or 1104	. . 3 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜃))
3	2	imbi1i 352	. 2 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓))
4		df-3an 1105	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) ↔ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓)) ∧ (𝜃 → 𝜓)))
5	1, 3, 4	3bitr4i 306	1 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   ∨ w3o 1102   ∧ 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-or 862  df-3or 1104  df-3an 1105

This theorem is used by:  3jaoi  1454