Theorem 3oran 1108

Description: Triple disjunction in terms of triple conjunction. (Contributed by NM, 8-Oct-2012.)

Assertion

Ref	Expression
3oran	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒))

Proof of Theorem 3oran

Step	Hyp	Ref	Expression
1		3ioran 1105	. . 3 ⊢ (¬ (𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒))
2	1	con1bii 356	. 2 ⊢ (¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒) ↔ (𝜑 ∨ 𝜓 ∨ 𝜒))
3	2	bicomi 224	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ¬ (¬ 𝜑 ∧ ¬ 𝜓 ∧ ¬ 𝜒))

Syntax hints:  ¬ wn 3   ↔ wb 206   ∨ w3o 1085   ∧ w3a 1086

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 207  df-an 396  df-or 848  df-3or 1087  df-3an 1088

This theorem is referenced by:  nolt02o  27613  nogt01o  27614  nosupbnd1lem6  27631  noinfbnd1lem6  27646  dalawlem10  39869