Theorem df3or2 44810

Description: Express triple-or in terms of implication and negation. Statement in [Frege1879] p. 11. (Contributed by RP, 25-Jul-2020.)

Assertion

Ref	Expression
df3or2	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (¬ 𝜑 → (¬ 𝜓 → 𝜒)))

Proof of Theorem df3or2

Step	Hyp	Ref	Expression
1		df-3or 1104	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
2		df-or 862	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (¬ (𝜑 ∨ 𝜓) → 𝜒))
3		ioran 999	. . . . 5 ⊢ (¬ (𝜑 ∨ 𝜓) ↔ (¬ 𝜑 ∧ ¬ 𝜓))
4	3	imbi1i 352	. . . 4 ⊢ ((¬ (𝜑 ∨ 𝜓) → 𝜒) ↔ ((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒))
5		impexp 456	. . . 4 ⊢ (((¬ 𝜑 ∧ ¬ 𝜓) → 𝜒) ↔ (¬ 𝜑 → (¬ 𝜓 → 𝜒)))
6	4, 5	bitri 278	. . 3 ⊢ ((¬ (𝜑 ∨ 𝜓) → 𝜒) ↔ (¬ 𝜑 → (¬ 𝜓 → 𝜒)))
7	2, 6	bitri 278	. 2 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) ↔ (¬ 𝜑 → (¬ 𝜓 → 𝜒)))
8	1, 7	bitri 278	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (¬ 𝜑 → (¬ 𝜓 → 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   ∨ w3o 1102

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  df-3or 1104

This theorem is used by: (None)