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DeMorgan’s Laws ( p  q )  ( p )  ( q ) ( p  q )  ( p )  ( q )

p q p  q ( p  q )  p  q ( p )  ( q ) 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1

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DeMorgan’s Laws ( p  q )  ( p )  ( q ) ( p  q )  ( p )  ( q )

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  1. p q pq (pq) p q(p) (q) 1 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 1 0 1 0 0 0 1 1 1 1 DeMorgan’s Laws ( p  q )  (p) (q) (p  q )  (p) (q) (Peter is tall and fat) Peter is not tall Peter is not fat (cucumbers are green or seedy)  cucumbers are not green cucumbers are not seedy

  2. p q pqp p  q • 1 1 1 0 1 • 1 0 0 0 0 • 0 1 1 1 1 • 0 0 1 1 1 Other important logical equivalences pq   ( p  q) (proof by contradiction)   p  q

  3. What is negation of implication?  ( p  q) ( p  q)  p (q)

  4. pq  q p   pq  p  q converse of pq inverse of pq • p q pq q  p • 1 1 1 1 • 1 0 0 1 • 0 1 1 0 • 0 0 1 1 p q p  q 0 0 1 0 1 1 1 0 0 1 1 1 An integer is divisible by 4  it is divisible by 2. An integer is divisible by 2  it is divisible by 4. An integer is not divisible by 4 it is not divisible by 2.

  5. contrapositive ofpq pq  ( q) ( p) Contrapositive law • p q pqq p( q) ( p) • 1 1 1 0 0 1 • 1 0 0 1 0 0 • 0 1 1 0 1 1 • 0 0 1 1 1 1 An integer is no divisible by 2  it is not divisible by 4.

  6. Double negation:p p p q = p  q = p 1 0 1 0 1 0

  7. a  b = b  a a +b = b + a a  (b  c)= (a  b) c a + (b + c) = (a + b) +c • Commutativity • p  q  q  p • p  q  q  p • Associativity • p (q  r)  (p  q )  r • p (q  r)  (p  q )  r

  8. p q r p  q q  r p (q  r) (p  q )  r 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 p (q  r)  (p  q )  r / ? p (q  r)  (p  q )  r

  9. Distributivity • p (q  r)  (p  q )  ( p  r) • p (q  r)  (p  q )  ( p  r) a (b+c) = a  b +a  c a +b  c = (a + b)  (a +c) Analogy with algebra is not complete!

  10. Identity • p  F  p p T  p • Domination • p  T  Tp F  F • Inverse • p p F p  p T • Idempotent • p p p p p p

  11. Absorption laws • p (p  q)  p • p  (p  q)  p q = T p (p  q)  p (p  T)  p p  p q = F p (p  q)  p (p  F)  p F p

  12. Example: Use the laws of logic to show that the following expression is a tautology [(pq)  (q r)]  [p (q  r)] Take the left-hand side and perform equivalent transformations: (pq)  (qr)   (pq) (qr))………………...equivalence  ( p q) (q  r)……………….DeMorgan’s law  ( ( p q) q ) r)…………….associative law  ( ( pq)  ( q q))  r)………distributive law  ( (pq )  T) r)………………inverse law  ( pq ) r)……………………...identity law  (p(q r))……………………..associative law p (qr)

  13. Deduction Rules (Inference rules) Suppose H1 H2 ...  HnC is a tautology, whereH1, H2, ... Hn are hypotheses and C is a conclusion. Then, given that all hypotheses are true, the conclusion is always true, or it is a valid argument. H1 H2 … Hn  C H1 H2 ...  HnC is called an inference rule.

  14. p pq  q • Examples: • [ p and (pq)]  q is a tautology (check in truth table) Modus Pones: Given that p and pq are both true Conclude: q

  15. pq q  p pq q r  pr 2. [(pq)  q]  p is a tautology, which leads to: Modus Tolens(proof by contradiction) Given: pq, q Conclude: p 3. Syllogism: Given: pq, q r Conclude: pr

  16. p  F  p 4. [p  F]p is a tautology Rule of Contradiction Given: p  F Conclude: p

  17. Predicates : p(x): x is a prime number. q(x): x > 2 r(x): x is an odd number These predicates are not propositions, because they can be true of false depending on x (unbound variable). p(2) is true, but p(4) is false q(3) is true, but q(1) is false

  18. Quantifiers. Universal quantifier xp(x) "for all x p(x) is true" "for any (every) xp(x) is true" For any x 2x is even. (universe of discourse is all integers ). When the domain (universe of discourse) is finite, xp(x) is equivalent to p(0)p(1)p(2)…p(n). All students in this class are CS majors.

  19. Existential quantifier xp(x) "there exists (at least one) x such that p(x) is true" " for some xp(x) is true" There exists a student in this class who likes discrete mathematics. xp(x). In this case the universe consists of students in this class and p(x) is the proposition "Student x likes discrete mathematics". xp(x) p(x1)p(x2)…p(x70)

  20. Proposition true false x p(x) p(x) is true for every x There is an x for which p(x) is false x p(x) There is an x for which p(x) is false for every x p(x) is false

  21. Negating Quantifiers Find negations of statements including quantifiers. (xp(x)) (All books are interesting) = There exists at least one book that is not interesting (xp(x)) (Some people like mathematics.) = Everybody dislikes mathematics (x p(x))   x p(x) ( x p(x))  x p(x)

  22.  x[ p(x)q(x) ] / Compound statements with existential quantifier x[ p(x)q(x) ]  [xp(x)][xq(x) ] [xp(x)][xq(x) ] x[ p(x) q(x) ]  [xp(x)] [xq(x) ] [xp(x)] [xq(x) ]  x[ p(x) q(x) ]

  23.  [xp(x)]  [xq(x) ] / Compound statements with universal quantifier x[ p(x)q(x) ]  [xp(x)][xq(x) ] [xp(x)][xq(x) ]  x[ p(x)q(x) ] x[ p(x)q(x) ] [xp(x)]  [xq(x) ]  x[ p(x)q(x) ]

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