Theorem 3jaodan 1453

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 416	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaodan.2	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → 𝜒)
4	3	ex 416	. . 3 ⊢ (𝜑 → (𝜃 → 𝜒))
5		3jaodan.3	. . . 4 ⊢ ((𝜑 ∧ 𝜏) → 𝜒)
6	5	ex 416	. . 3 ⊢ (𝜑 → (𝜏 → 𝜒))
7	2, 4, 6	3jaod 1451	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
8	7	imp 410	1 ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒)

Syntax hints:   → wi 4   ∧ wa 399   ∨ w3o 1098

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 209  df-an 400  df-or 859  df-3or 1100  df-3an 1101

This theorem is referenced by:  mpjao3dan  1454  onzsl  7828  zeo  12661  xrltnsym  13141  xrlttri  13143  xrlttr  13144  qbtwnxr  13205  xltnegi  13221  xaddcom  13245  xnegdi  13253  xsubge0  13266  xrub  13317  bpoly3  16090  blssioo  24857  ismbf2d  25704  itg2seq  25806  eliccioo  33110  3ccased  36074  lineelsb2  36503  sticksstones1  42768  dfxlim2v  46426  usgrexmpl2trifr  48664