Theorem 3jaodan 1458

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 1456	. 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  1459  onzsl  7857  zeo  12785  xrltnsym  13266  xrlttri  13268  xrlttr  13269  qbtwnxr  13330  xltnegi  13346  xaddcom  13370  xnegdi  13378  xsubge0  13391  xrub  13442  bpoly3  16224  blssioo  25114  ismbf2d  25961  itg2seq  26063  eliccioo  33497  3ccased  36484  lineelsb2  36913  sticksstones1  43197  dfxlim2v  46857  usgrexmpl2trifr  49135