Theorem 3jaob 1422

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

Assertion

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

Proof of Theorem 3jaob

Step	Hyp	Ref	Expression
1		3mix1 1326	. . . 4 ⊢ (𝜑 → (𝜑 ∨ 𝜒 ∨ 𝜃))
2	1	imim1i 63	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜑 → 𝜓))
3		3mix2 1327	. . . 4 ⊢ (𝜒 → (𝜑 ∨ 𝜒 ∨ 𝜃))
4	3	imim1i 63	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜒 → 𝜓))
5		3mix3 1328	. . . 4 ⊢ (𝜃 → (𝜑 ∨ 𝜒 ∨ 𝜃))
6	5	imim1i 63	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜃 → 𝜓))
7	2, 4, 6	3jca 1124	. 2 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))
8		3jao 1421	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))
9	7, 8	impbii 211	1 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))

Syntax hints:   → wi 4   ↔ wb 208   ∨ w3o 1082   ∧ w3a 1083

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 209  df-an 399  df-or 844  df-3or 1084  df-3an 1085

This theorem is referenced by: (None)