Theorem 3jaoian 1300

Description: Disjunction of 3 antecedents (inference). (Contributed by NM, 14-Oct-2005.)

Hypotheses

Ref	Expression
3jaoian.1	⊢ ((𝜑 ∧ 𝜓) → 𝜒)
3jaoian.2	⊢ ((𝜃 ∧ 𝜓) → 𝜒)
3jaoian.3	⊢ ((𝜏 ∧ 𝜓) → 𝜒)

Assertion

Ref	Expression
3jaoian	⊢ (((𝜑 ∨ 𝜃 ∨ 𝜏) ∧ 𝜓) → 𝜒)

Proof of Theorem 3jaoian

Step	Hyp	Ref	Expression
1		3jaoian.1	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜒)
2	1	ex 114	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3		3jaoian.2	. . . 4 ⊢ ((𝜃 ∧ 𝜓) → 𝜒)
4	3	ex 114	. . 3 ⊢ (𝜃 → (𝜓 → 𝜒))
5		3jaoian.3	. . . 4 ⊢ ((𝜏 ∧ 𝜓) → 𝜒)
6	5	ex 114	. . 3 ⊢ (𝜏 → (𝜓 → 𝜒))
7	2, 4, 6	3jaoi 1298	. 2 ⊢ ((𝜑 ∨ 𝜃 ∨ 𝜏) → (𝜓 → 𝜒))
8	7	imp 123	1 ⊢ (((𝜑 ∨ 𝜃 ∨ 𝜏) ∧ 𝜓) → 𝜒)

Syntax hints:   → wi 4   ∧ wa 103   ∨ w3o 972

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 704

This theorem depends on definitions:  df-bi 116  df-3or 974  df-3an 975

This theorem is referenced by:  xrltnsym  9750  xrlttr  9752  xltnegi  9792  xaddcom  9818  xnegdi  9825  qbtwnxr  10214