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

Mathematics for Comter I 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. Classic Example: P  P A contingency is a proposition which neither a tautology nor a contradiction. Example: (P  Q)  R

Mathematics for Comter I 3.1. Propositional Equivalences Two propositions P and Q are logically equivalent if P  Q is a tautology. We write P  Q

Mathematics for Comter I 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 example or to disprove the assertion. search for a case where the proposition is false.

Mathematics for Comter I 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  P need 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) PQ PQPQ PQ PQPQ

Mathematics for Comter I 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) PQ PQPQ PQ PQPQ

Mathematics for Comter I 3.1. Propositional Equivalences Some important logical equivalences: EquivalencesName P  T  P P  F  P Identity laws P  T  T P  F  F Domination laws P  P  P P  P  P Idempotent laws  (  P)  P Double negation law PT PTPT PP PPPP

Mathematics for Comter I 3.1. Propositional Equivalences Some important logical equivalences: EquivalencesName P  Q  Q  P P  Q  Q  P Commutative laws (P  Q)  R  P  (Q  R) (P  Q)  R  P  (Q  R) Associative laws P  (Q  R)  (P  Q)  (P  R) P  (Q  R)  (P  Q)  (P  R) Distributive laws  (P  Q)   P  Q  (P  Q)   P  Q De Morgan’s laws

Mathematics for Comter I 3.1. Propositional Equivalences Other logical equivalences: EquivalencesName P  Q   P  Q Implication P  P  T Tautology P  P  F Contradiction (P  Q)  (Q  P)  (P  Q) Equivalence (P  Q)  (P  Q)   P Absurdity

Mathematics for Comter I 3.1. Propositional Equivalences Other logical equivalences: Equivalent expressions can always be substituted for each other in a more complex expression – useful for simplification. EquivalencesName (P  Q)  (  Q  P) Contrapositive P  (P  Q)  P P  (P  Q)  P Absorption (P  Q)  R  P  (Q  R) Exportation

Mathematics for Comter I 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  Qfrom the identity law for F   (P  Q) from the second De Morgan’s law

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

Mathematics for Comter I 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

Mathematics for Comter I 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.

Mathematics for Comter I 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)

Mathematics for Comter I 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

Mathematics for Comter I Quantifiers Universal P(x) is true for every x in the universe of discourse. Notation: universal quantifier  x P(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)

Mathematics for Comter I Quantifiers Existential P(x) is true for some x in the universe of discourse. Notation: existential quantifier  x P(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)

Mathematics for Comter I Quantifiers 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. REMEMBER! A predicate (propositional function) is not a proposition until all variables have been bound either by quantification or assignment of a value!

Mathematics for Comter I Quantifiers Helpful hints: To memorize the predicate equivalences – think about searching and looping: Assume there are n objects in U. To determine whether  x P(x) is true: Loop through all n values of x to see if P(x) is always true. If there is a value x for which P(x) is false, then we have shown that  x P(x) is false, i.e. we have shown  x P(x) or  x  P(x)

Mathematics for Comter I Quantifiers To determine whether  x P(x) is true: Loop through all n values of x searching for a value for which P(x) is true. If we find one, then  x P(x) is true. Otherwise  x P(x) is false, i.e.  x P(x) or  x  P(x)

Mathematics for Comter I 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?

Mathematics for Comter I 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!

Mathematics for Comter I 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.

Mathematics for Comter I 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))

Mathematics for Comter I 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)))

Mathematics for Comter I 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.

Mathematics for Comter I 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))?

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