Theorem 3jaob

Description: Disjunction of three antecedents. (Contributed by NM, 13-Sep-2011)

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

Proof

Step	Hyp	Ref	Expression
1		3mix1	⊢ ( 𝜑 → ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) )
2	1	imim1i	⊢ ( ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) → ( 𝜑 → 𝜓 ) )
3		3mix2	⊢ ( 𝜒 → ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) )
4	3	imim1i	⊢ ( ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) → ( 𝜒 → 𝜓 ) )
5		3mix3	⊢ ( 𝜃 → ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) )
6	5	imim1i	⊢ ( ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) → ( 𝜃 → 𝜓 ) )
7	2 4 6	3jca	⊢ ( ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) → ( ( 𝜑 → 𝜓 ) ∧ ( 𝜒 → 𝜓 ) ∧ ( 𝜃 → 𝜓 ) ) )
8		3jao	⊢ ( ( ( 𝜑 → 𝜓 ) ∧ ( 𝜒 → 𝜓 ) ∧ ( 𝜃 → 𝜓 ) ) → ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) )
9	7 8	impbii	⊢ ( ( ( 𝜑 ∨ 𝜒 ∨ 𝜃 ) → 𝜓 ) ↔ ( ( 𝜑 → 𝜓 ) ∧ ( 𝜒 → 𝜓 ) ∧ ( 𝜃 → 𝜓 ) ) )