Theorem 3jaob

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

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

Proof

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