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Lecture 3: Logic (2)

Lecture 3: Logic (2). Propositional Equivalences Predicates and Quantifiers. 3.1. Propositional Equivalences. Categories of compound propositions: A tautology is a proposition which is always true . Classic Example: P  P A contradiction is a proposition which is always false .

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Lecture 3: Logic (2)

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  1. Lecture 3: Logic (2) • Propositional Equivalences • Predicates and Quantifiers 310205 Mathematics for Comter I

  2. 3.1. Propositional Equivalences • Categories of compound propositions: • A tautology is a proposition which is always true. • Classic Example: PP • A contradictionis a proposition which is always false. • Classic Example: PP • A contingencyis a proposition which neither a tautology nor a contradiction. • Example: (PQ)R 310205 Mathematics for Comter I

  3. 3.1. Propositional Equivalences • Two propositions P and Q are logically equivalentif P  Q is a tautology. We write P  Q 310205 Mathematics for Comter I

  4. 3.1. Propositional Equivalences • Example: (PQ)(QP) (P  Q) • Proof: • Left side and the right side must have the same truth values, independent of the truth value of the component propositions. • To show a proposition is not a tautology: use an abbreviated truth table • try to find a counter exampleor to disprove the assertion. • search for a case where the proposition is false. 310205 Mathematics for Comter I

  5. 3.1. Propositional Equivalences • Two possible cases: • Case 1: Left side false, right side true. • Case 2: Left side true, right side false. • Case 1: Try left side false, right side true • Left side false: only one of PQ or QPneed be false. 1a. Assume PQ= F. Then P = T, Q = F. But then right side PQ = F. Oops, wrong guess. 1b. Try QP= F. Then Q = T, P = F. But then right side PQ = F. Another wrong guess. *Proof for (PQ)(QP)  (P  Q) 310205 Mathematics for Comter I

  6. 3.1. Propositional Equivalences • Case 2. Try left side true, right side false • If right side is false, P and Q cannot have the same truth value. 2a. Assume P =T, Q = F. Then PQ= F and the conjunction must be false so the left side cannot be true in this case. Another wrong guess. 2b. Assume Q = T, P = F. Then QP= F. Again the left side cannot be true. • We have exhausted all possibilities and not found a counter-example. The two propositions must be logically equivalent. • Note: Given such equivalence, if and only if or iff is also stated as is a necessary and sufficient condition for. *Proof for (PQ)(QP) (P  Q) 310205 Mathematics for Comter I

  7. 3.1. Propositional Equivalences • Some important logical equivalences: 310205 Mathematics for Comter I

  8. 3.1. Propositional Equivalences • Some important logical equivalences: 310205 Mathematics for Comter I

  9. 3.1. Propositional Equivalences • Other logical equivalences: 310205 Mathematics for Comter I

  10. 3.1. Propositional Equivalences • Other logical equivalences: • Equivalent expressions can always be substituted for each other in a more complex expression – useful for simplification. 310205 Mathematics for Comter I

  11. 3.1. Propositional Equivalences • Example: • (P(PQ)) can be simplified by using the following series of logical equivalence: (P(PQ)) P(PQ)) from the second De Morgan’s law P[(P)Q] from the first De Morgan’s law P(PQ) from the double negation law (PP)(PQ] from the distributive law  F(PQ) since PPF PQ from the identity law for F (PQ) from the second De Morgan’s law 310205 Mathematics for Comter I

  12. 3.1. Propositional Equivalences • But Complexity (2n)… REMEMBER! We can always use a truth table to show that the simplified proposition is equivalent to the original proposition. 310205 Mathematics for Comter I

  13. 3.2. Predicates and Quantifiers • 2+1=3: a proposition • x+y=3: propositional functions or predicates • A generalization of propositions • Propositions which contain variables • Predicates become propositions once every variable is bound - by • assigning it a value from the Universe of Discourse U or • quantifying it 310205 Mathematics for Comter I

  14. 3.2.1. Predicates • Example 1: • Let U = Z, the integers = , -2, -1, 0 , 1, 2, 3,  • P(x): x > 0,a predicate or propositional function. • It has no truth value until the variable x is bound. • Examples of propositions where x is assigned a value: • P(-3) is false, i.e. -3 > 0 is false. • P(0) is false. • P(3) is true. • The collection of integers for which P(x) is true are the positive integers. 310205 Mathematics for Comter I

  15. 3.2.1. Predicates • Example 1 (continued): • P(x): x > 0 is the predicate. • P(y)P(0) is not a proposition. The variable y has not been bound. • However, P(3)P(0) is a proposition which is true. • Example 2: • Let R be the three-variable predicate R(x, y, z): x + y = z. Find the truth value of • R(2, -1, 5) • R(3, 4, 7) • R(x, 3, z) 310205 Mathematics for Comter I

  16. 3.2.2. Quantifiers • Specific value vs. Range • What range of values in U for which the bounded propositions are true? • Two possibilities: • Universal: For all values in U • Existential: For some values in U 310205 Mathematics for Comter I

  17. 3.2.2. Quantifiers • Universal • P(x) is true for every x in the universe of discourse. • Notation: universal quantifier xP(x) • For all x, P(x) or • For every x, P(x) • The variable x is bound by the universal quantifier producing a proposition. • Example: • U = { 1,2,3 } x P(x) P(1)P(2)P(3) 310205 Mathematics for Comter I

  18. 3.2.2. Quantifiers • Existential • P(x) is true for some x in the universe of discourse. • Notation: existential quantifierxP(x) • There is an x such that P(x) or • For some x, P(x) • For at least one x, P(x) • I can find an x such that P(x) • Example: • U = { 1,2,3 } x P(x)P(1)P(2)P(3) 310205 Mathematics for Comter I

  19. 3.2.2. Quantifiers REMEMBER! A predicate (propositional function) is not a proposition until all variables have been bound either by quantification or assignment of a value! • Predicate equivalences: • Equivalences involving the negation operator • x P(x)  x P(x) • x P(x)  x P(x) • Distributing a negation operator across a quantifier changes a universal to an existential, and vice versa. 310205 Mathematics for Comter I

  20. 3.2.2. Quantifiers • Multiple Quantifiers: • Read from left to right . . . • Example 1: • Let U = R, the real numbers, P(x,y): xy = 0 xy P(x,y) xy P(x,y) xy P(x,y) xy P(x,y) • The only one that is false is the first one. Why? • Suppose P(x,y) is the predicate x/y=1? 310205 Mathematics for Comter I

  21. 3.2.2. Quantifiers • Dangerous situations: • Commutativity of quantifiers xy P(x, y) yx P(x, y)? YES! xy P(x, y) yx P(x, y)? NO! DIFFERENT MEANING! 310205 Mathematics for Comter I

  22. 3.2.2. Quantifiers • Example of non-commutativity of quantifiers: • Let Q(x, y) denote “x + y = 0.” • Are the truth values of the quantifications yx P(x, y) and xy P(x, y) the same? • The answer is NO since: • yx P(x, y) means “There is a real number y such that for all real numbers x, Q(x, y) is true.” • The statement is false. Why? • x y P(x, y) means “For every real number x there is a real number y such that Q(x, y) is true.” • The statement is true. 310205 Mathematics for Comter I

  23. 3.2.3. Converting from English • (can be very difficult) • Example 1: • Express the statement “If somebody is female and is a parent, then this person is someone’s mother” as a logical expression. • Let F(x): x is female. P(x): x is a parent M(x, y): x is the mother of y. • The statement applies to all people. x ((F(x)  (P(x))  y M(x, y)) 310205 Mathematics for Comter I

  24. 3.2.3. Converting from English • Example 2: • Express the statement “Everyone has exactly one best friend” as a logical expression. • Let B(x, y): y is the best friend of x. • The statement says “exactly one best friend”. This means that if y is the best friend of x, then all other people z other than y can not be the best friend of x. xyz (B(x, y)  ((z  y)  B(x, z))) 310205 Mathematics for Comter I

  25. 3.2.3. Converting from English • Example 3: • Consider the following statements. “All lions are fierce.” “Some lions do not drink coffee.” “Some fierce creatures do not drink coffee.” • The first two are called premises and the third is called the conclusion. The entire set is called an argument. 310205 Mathematics for Comter I

  26. 3.2.3. Converting from English • Example 3 (continued): • We can express these statements as follows. • LetP(x): x is a lion. Q(x): x is fierce. R(x): x drinks coffee. • Then x (P(x)  Q(x)). x (P(x)  R(x)). x (Q(x)  R(x)). • Why can’t we write the second statement as x (P(x)  R(x))? 310205 Mathematics for Comter I

  27. 3.3. Further Readings • Propositional Equivalences • Rosen: Section 1.2. • Predicates and Quantifiers • Rosen: Section 1.3. 310205 Mathematics for Comter I

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