Theorem 3jao 1296

Description: Disjunction of 3 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		df-3or 974	. 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜃))
2		jao 750	. . . 4 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜑 ∨ 𝜒) → 𝜓)))
3		jao 750	. . . 4 ⊢ (((𝜑 ∨ 𝜒) → 𝜓) → ((𝜃 → 𝜓) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓)))
4	2, 3	syl6 33	. . 3 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜃 → 𝜓) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓))))
5	4	3imp 1188	. 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓))
6	1, 5	syl5bi 151	1 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓))

Syntax hints:   → wi 4   ∨ wo 703   ∨ w3o 972   ∧ w3a 973

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704

This theorem depends on definitions:  df-bi 116  df-3or 974  df-3an 975

This theorem is referenced by:  3jaob  1297  3jaoi  1298  3jaod  1299