Theorem 3jaodan 1317

Description: Disjunction of 3 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 115	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaodan.2	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → 𝜒)
4	3	ex 115	. . 3 ⊢ (𝜑 → (𝜃 → 𝜒))
5		3jaodan.3	. . . 4 ⊢ ((𝜑 ∧ 𝜏) → 𝜒)
6	5	ex 115	. . 3 ⊢ (𝜑 → (𝜏 → 𝜒))
7	2, 4, 6	3jaod 1315	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
8	7	imp 124	1 ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒)

Syntax hints:   → wi 4   ∧ wa 104   ∨ w3o 979

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710

This theorem depends on definitions:  df-bi 117  df-3or 981  df-3an 982

This theorem is referenced by:  zeo  9431  xrltnsym  9868  xrlttr  9870  xrltso  9871  xrlttri3  9872  xltnegi  9910  xaddcom  9936  xnegdi  9943  xsubge0  9956  qbtwnxr  10347  blssioo  14789