Theorem jaodd 43035

Description: Double deduction form of jaoi 871. (Contributed by Steven Nguyen, 17-Jul-2022.)

Hypotheses

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

Assertion

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

Proof of Theorem jaodd

Step	Hyp	Ref	Expression
1		jaodd.1	. 2 ⊢ (𝜑 → (𝜓 → (𝜒 → 𝜃)))
2		jaodd.2	. 2 ⊢ (𝜑 → (𝜓 → (𝜏 → 𝜃)))
3		jao 975	. 2 ⊢ ((𝜒 → 𝜃) → ((𝜏 → 𝜃) → ((𝜒 ∨ 𝜏) → 𝜃)))
4	1, 2, 3	syl6c 71	1 ⊢ (𝜑 → (𝜓 → ((𝜒 ∨ 𝜏) → 𝜃)))

This proof depends on syntax axioms:   → wi 4   ∨ wo 861

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862

This theorem is used by: (None)