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Revision: Negation of propositional formulae Conjunctive and - - PowerPoint PPT Presentation

Revision: Negation of propositional formulae Conjunctive and disjunctive normal forms of propositional formulae Valentin Goranko DTU Informatics September 2010 V Goranko Important equivalences for negations of propositional formulae


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V Goranko

Revision: Negation of propositional formulae Conjunctive and disjunctive normal forms of propositional formulae

Valentin Goranko DTU Informatics September 2010

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Important equivalences for negations of propositional formulae

  • ¬¬A ≡ A,
  • ¬(A ∧ B) ≡ ¬A ∨ ¬B,
  • ¬(A ∨ B) ≡ ¬A ∧ ¬B,
  • ¬(A → B) ≡ A ∧ ¬B,
  • ¬(A ↔ B)

≡ ¬((A → B) ∧ (B → A)) ≡ ¬(A → B) ∨ ¬(B → A) ≡ (A ∧ ¬B) ∨ (B ∧ ¬A).

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Negating propositional formulae: example

¬(¬B → (¬C ∧ D)) ≡ ¬B ∧ ¬(¬C ∧ D) ≡ ¬B ∧ (¬¬C ∨ ¬D) ≡ ¬B ∧ (C ∨ ¬D).

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Normal forms: basic definitions

  • 1. A literal is a propositional variable or its negation.
  • 2. An elementary disjunction (respectively, elementary

conjunction) is a disjunction (respectively, conjunction) of

  • literals. Examples:

p, ¬q, p ∨ ¬q, p ∨ ¬p ∨ q ∨ ¬r are elementary disjunctions; p, ¬q, ¬p ∧ q, ¬p ∧ q ∧ ¬r ∧ ¬p are elementary conjunctions.

  • 3. A disjunctive normal form (DNF) is a disjunction of

elementary conjunctions. Examples: p, ¬q, p ∧ ¬q, p ∨ ¬q, (p ∧ ¬p) ∨ ¬q, (r ∧ q ∧ ¬p) ∨ (¬q ∧ p) ∨ (¬r ∧ p).

  • 4. A conjunctive normal form (CNF) is a conjunction of

elementary disjunctions. Examples: p, ¬q, p ∧ ¬q, p ∨ ¬q, p ∧ (¬p ∨ ¬q), (r ∨ q ∨ ¬r) ∧ ¬q ∧ (¬p ∨ r).

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Algorithm for equivalent transformation to CNF/DNF

Theorem: Every propositional formula is equivalent to a disjunctive normal form and to a conjunctive normal form. Algorithm transforming a formula into a DNF, respectively CNF:

  • 1. Eliminate all occurrences of ↔ and → using the logical

equivalences A → B ≡ ¬A ∨ B, A ↔ B ≡ (A → B) ∧ (B → A).

  • 2. Import all negations in front of the propositional variables,

using the logical equivalences listed above.

  • 3. For a DNF: distribute all conjunctions over disjunctions using

p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

  • 4. Respectively, for a CNF: distribute all disjunctions over

conjunctions using p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

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Some useful simplifications

Throughout this process the formulae can be simplified by using commutativity, associativity, and idempotency of ∨ and, as well as: ◮ p ∨ ¬p ≡ ⊤; p ∧ ¬p ≡ ⊥; ◮ p ∧ ⊤ ≡ p; p ∧ ⊥ ≡ ⊥; ◮ p ∨ ⊤ ≡ ⊤; p ∨ ⊥ ≡ p.

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Example

(p ∧ ¬r) → (p ↔ ¬q) ≡ ¬(p ∧ ¬r) ∨ ((p → ¬q) ∧ (¬q → p)) ≡ (¬p ∨ ¬¬r) ∨ ((¬p ∨ ¬q) ∧ (¬¬q ∨ p)) ≡ ¬p ∨ r ∨ ((¬p ∨ ¬q) ∧ (q ∨ p)) For a DNF we further distribute ∧ over ∨ and simplify: ≡ ¬p ∨ r ∨ (((¬p ∨ ¬q) ∧ q) ∨ ((¬p ∨ ¬q) ∧ p)) ≡ ¬p ∨ r ∨ ((¬p ∧ q) ∨ (¬q ∧ q)) ∨ ((¬p ∧ p) ∨ (¬q ∧ p)) ≡ ¬p ∨ r ∨ ((¬p ∧ q) ∨ ⊥) ∨ (⊥ ∨ (¬q ∧ p)) ≡ ¬p ∨ r ∨ (¬p ∧ q) ∨ (¬q ∧ p). For a CNF we distribute ∨ over ∧ and simplify: ≡ (¬p ∨ r ∨ ¬p ∨ ¬q) ∧ (¬p ∨ r ∨ q ∨ p) ≡ (¬p ∨ r ∨ ¬q) ∧ (⊤ ∨ r ∨ q) ≡ (¬p ∨ r ∨ ¬q) ∧ ⊤ ≡ ¬p ∨ r ∨ ¬q.