Theorem 3jaodan 1296

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 114	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaodan.2	. . . 4 ⊢ ((𝜑 ∧ 𝜃) → 𝜒)
4	3	ex 114	. . 3 ⊢ (𝜑 → (𝜃 → 𝜒))
5		3jaodan.3	. . . 4 ⊢ ((𝜑 ∧ 𝜏) → 𝜒)
6	5	ex 114	. . 3 ⊢ (𝜑 → (𝜏 → 𝜒))
7	2, 4, 6	3jaod 1294	. 2 ⊢ (𝜑 → ((𝜓 ∨ 𝜃 ∨ 𝜏) → 𝜒))
8	7	imp 123	1 ⊢ ((𝜑 ∧ (𝜓 ∨ 𝜃 ∨ 𝜏)) → 𝜒)

Syntax hints:   → wi 4   ∧ wa 103   ∨ w3o 967

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699

This theorem depends on definitions:  df-bi 116  df-3or 969  df-3an 970

This theorem is referenced by:  zeo  9296  xrltnsym  9729  xrlttr  9731  xrltso  9732  xrlttri3  9733  xltnegi  9771  xaddcom  9797  xnegdi  9804  xsubge0  9817  qbtwnxr  10193  blssioo  13185