Theorem 3jao 1422

Description: Disjunction of three antecedents. (Contributed by NM, 8-Apr-1994.)

Assertion

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

Proof of Theorem 3jao

Step	Hyp	Ref	Expression
1		jao 958	. . 3 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜑 ∨ 𝜒) → 𝜓)))
2		df-3or 1085	. . . 4 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜃))
3		jao 958	. . . 4 ⊢ (((𝜑 ∨ 𝜒) → 𝜓) → ((𝜃 → 𝜓) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓)))
4	2, 3	syl7bi 258	. . 3 ⊢ (((𝜑 ∨ 𝜒) → 𝜓) → ((𝜃 → 𝜓) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)))
5	1, 4	syl6 35	. 2 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜃 → 𝜓) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))))
6	5	3imp 1108	1 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))

Syntax hints:   → wi 4   ∨ wo 844   ∨ w3o 1083   ∧ w3a 1084

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 210  df-an 400  df-or 845  df-3or 1085  df-3an 1086

This theorem is referenced by:  3jaob  1423  3jaoi  1424  3jaod  1425