Theorem 3jaodd 33559

Description: Double deduction form of 3jaoi 1425. (Contributed by Scott Fenton, 20-Apr-2011.)

Hypotheses

Ref	Expression
3jaodd.1	⊢ (𝜑 → (𝜓 → (𝜒 → 𝜂)))
3jaodd.2	⊢ (𝜑 → (𝜓 → (𝜃 → 𝜂)))
3jaodd.3	⊢ (𝜑 → (𝜓 → (𝜏 → 𝜂)))

Assertion

Ref	Expression
3jaodd	⊢ (𝜑 → (𝜓 → ((𝜒 ∨ 𝜃 ∨ 𝜏) → 𝜂)))

Proof of Theorem 3jaodd

Step	Hyp	Ref	Expression
1		3jaodd.1	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜒 → 𝜂)))
2	1	com3r 87	. . 3 ⊢ (𝜒 → (𝜑 → (𝜓 → 𝜂)))
3		3jaodd.2	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜂)))
4	3	com3r 87	. . 3 ⊢ (𝜃 → (𝜑 → (𝜓 → 𝜂)))
5		3jaodd.3	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜏 → 𝜂)))
6	5	com3r 87	. . 3 ⊢ (𝜏 → (𝜑 → (𝜓 → 𝜂)))
7	2, 4, 6	3jaoi 1425	. 2 ⊢ ((𝜒 ∨ 𝜃 ∨ 𝜏) → (𝜑 → (𝜓 → 𝜂)))
8	7	com3l 89	1 ⊢ (𝜑 → (𝜓 → ((𝜒 ∨ 𝜃 ∨ 𝜏) → 𝜂)))

Syntax hints:   → wi 4   ∨ w3o 1084

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 206  df-an 396  df-or 844  df-3or 1086  df-3an 1087

This theorem is referenced by: (None)