Theorem 3jaodan 1426

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 415	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaodan.2	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → 𝜒)
4	3	ex 415	. . 3 ⊢ (𝜑 → (𝜃 → 𝜒))
5		3jaodan.3	. . . 4 ⊢ ((𝜑 ∧ 𝜏) → 𝜒)
6	5	ex 415	. . 3 ⊢ (𝜑 → (𝜏 → 𝜒))
7	2, 4, 6	3jaod 1424	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
8	7	imp 409	1 ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒)

Syntax hints:   → wi 4   ∧ wa 398   ∨ w3o 1082

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 399  df-or 844  df-3or 1084  df-3an 1085

This theorem is referenced by:  mpjao3dan  1427  onzsl  7564  zeo  12071  xrltnsym  12533  xrlttri  12535  xrlttr  12536  qbtwnxr  12596  xltnegi  12612  xaddcom  12636  xnegdi  12644  xsubge0  12657  xrub  12708  bpoly3  15415  blssioo  23406  ismbf2d  24244  itg2seq  24346  eliccioo  30611  3ccased  32953  lineelsb2  33613  dfxlim2v  42134