Theorem 3jaodan 1459

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

Hypotheses

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

Assertion

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

Proof of Theorem 3jaodan

Step	Hyp	Ref	Expression
1		3jaodan.1	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
2	1	ex 418	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaodan.2	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → 𝜒)
4	3	ex 418	. . 3 ⊢ (𝜑 → (𝜃 → 𝜒))
5		3jaodan.3	. . . 4 ⊢ ((𝜑 ∧ 𝜏) → 𝜒)
6	5	ex 418	. . 3 ⊢ (𝜑 → (𝜏 → 𝜒))
7	2, 4, 6	3jaod 1457	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
8	7	imp 412	1 ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∨ 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 402  df-or 862  df-3or 1104  df-3an 1105

This theorem is used by:  mpjao3dan  1460  onzsl  7878  zeo  12809  xrltnsym  13290  xrlttri  13292  xrlttr  13293  qbtwnxr  13354  xltnegi  13370  xaddcom  13394  xnegdi  13402  xsubge0  13415  xrub  13466  bpoly3  16250  blssioo  25141  ismbf2d  25988  itg2seq  26090  eliccioo  33564  3ccased  36661  lineelsb2  37090  sticksstones1  43387  dfxlim2v  47060  usgrexmpl2trifr  49338