Theorem 3jaoian 1457

Description: Disjunction of three antecedents (inference). (Contributed by NM, 14-Oct-2005.)

Hypotheses

Ref	Expression
3jaoian.1	⊢ ((𝜑 ∧ 𝜓) → 𝜒)
3jaoian.2	⊢ ((𝜃 ∧ 𝜓) → 𝜒)
3jaoian.3	⊢ ((𝜏 ∧ 𝜓) → 𝜒)

Assertion

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

Proof of Theorem 3jaoian

Step	Hyp	Ref	Expression
1		3jaoian.1	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
2	1	ex 417	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaoian.2	. . . 4 ⊢ ((𝜃 ∧ 𝜓) → 𝜒)
4	3	ex 417	. . 3 ⊢ (𝜃 → (𝜓 → 𝜒))
5		3jaoian.3	. . . 4 ⊢ ((𝜏 ∧ 𝜓) → 𝜒)
6	5	ex 417	. . 3 ⊢ (𝜏 → (𝜓 → 𝜒))
7	2, 4, 6	3jaoi 1454	. 2 ⊢ ((𝜑 ∨ 𝜃 ∨ 𝜏) → (𝜓 → 𝜒))
8	7	imp 411	1 ⊢ (((𝜑 ∨ 𝜃 ∨ 𝜏) ∧ 𝜓) → 𝜒)

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   ∨ w3o 1102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3or 1104  df-3an 1105

This theorem is used by:  xrltnsym  13166  xrlttri  13168  xrlttr  13169  qbtwnxr  13230  xltnegi  13246  xaddcom  13270  xnegdi  13278  lcmftp  16698  xaddeq0  33107  3ccased  36219  usgrexmpl2trifr  48830