Theorem 3oran 951

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 950	. . 3 ⊢ (¬ (φ ∨ ψ ∨ χ) ↔ (¬ φ ∧ ¬ ψ ∧ ¬ χ))
2	1	con1bii 321	. 2 ⊢ (¬ (¬ φ ∧ ¬ ψ ∧ ¬ χ) ↔ (φ ∨ ψ ∨ χ))
3	2	bicomi 193	1 ⊢ ((φ ∨ ψ ∨ χ) ↔ ¬ (¬ φ ∧ ¬ ψ ∧ ¬ χ))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 176   ∨ w3o 933   ∧ w3a 934

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

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3or 935  df-3an 936

This theorem is used by: (None)