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lecture3 - Lecture 3 Predicate Logic predicates and...

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Lecture 3 Predicate Logic: predicates and quantifiers

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So far Propositional Logic Propositional Equivalences Truth tables Derivations using basic formulae
Recap p T p p F p Identity Laws (p q) r p (q r) (p q) r p (q r) Associative laws p T T p F F Domination Law p (q r) (p q) (p r) p (q r) (p q) (p r) Distributive laws p p p p p p Idempotent Laws ¬ (p q) ≡ ¬ p ∨ ¬ q ¬ (p q) ≡ ¬ p ∧ ¬ q De Morgan’s laws ¬ ( ¬ p) p Double negation law p (p q) p p (p q) p Absorption laws p q q p p q q p Commutative Laws p ∨ ¬ p T p ∧ ¬ p F Negation laws p q ≡ ¬ p q Definition of Implication p q (p q) (q p) Definition of Biconditional

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Example Show that [ ¬ p (p q)] q is a tautology
Solution 1 [ ¬ p (p q)] q ≡ ¬ [ ¬ p (p q)] q (implication) [p ∨ ¬ (p q)] q (de morgan) [p ( ¬ p ∧ ¬ q)] q (de morgan) [(p ∨ ¬ p) (p ∨¬ q)] q (distribution) [ T (p ∨ ¬ q)] q (negation) [ (p ∨ ¬ q)] q (identity) p ( ¬ q q) (associativity) p T T (negation, domination)

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Solution 2 We want to show that [ ¬ p (p q)] q is a tautology There are two cases: Case 1:Hypothesis is false. Then the implication is true. Case 2: Hypothesis is true. We must show that the conclusion is true. Case 2: [
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lecture3 - Lecture 3 Predicate Logic predicates and...

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