Theorem oran 255

Description: Disjunction in terms of conjunction (DeMorgan's law). Compare Theorem *4.57 of [WhiteheadRussell] p. 120.

Assertion

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

Proof of Theorem oran

Step	Hyp	Ref	Expression
1		pm4.13 142	. 2 ⊢ ((φ ∨ ψ) ↔ ¬ ¬ (φ ∨ ψ))
2		ioran 254	. . 3 ⊢ (¬ (φ ∨ ψ) ↔ (¬ φ ∧ ¬ ψ))
3	2	negbii 162	. 2 ⊢ (¬ ¬ (φ ∨ ψ) ↔ ¬ (¬ φ ∧ ¬ ψ))
4	1, 3	bitr 151	1 ⊢ ((φ ∨ ψ) ↔ ¬ (¬ φ ∧ ¬ ψ))

Syntax hints:  ¬ wn 1   ↔ wb 127   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  orim12i 271  jao 274  andi 456  19.43 767  dmsnsn0 2544  mdsym 5784

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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