Theorem 3jaob 1302

Description: Disjunction of 3 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 1166	. . . 4 ⊢ (𝜑 → (𝜑 ∨ 𝜒 ∨ 𝜃))
2	1	imim1i 60	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜑 → 𝜓))
3		3mix2 1167	. . . 4 ⊢ (𝜒 → (𝜑 ∨ 𝜒 ∨ 𝜃))
4	3	imim1i 60	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜒 → 𝜓))
5		3mix3 1168	. . . 4 ⊢ (𝜃 → (𝜑 ∨ 𝜒 ∨ 𝜃))
6	5	imim1i 60	. . 3 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → (𝜃 → 𝜓))
7	2, 4, 6	3jca 1177	. 2 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) → ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))
8		3jao 1301	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))
9	7, 8	impbii 126	1 ⊢ (((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)))

Syntax hints:   → wi 4   ↔ wb 105   ∨ w3o 977   ∧ w3a 978

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709

This theorem depends on definitions:  df-bi 117  df-3or 979  df-3an 980

This theorem is referenced by: (None)