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Mat h ematic al logi c

Mat h ematic al logi c. Lesson 5 Rela tions , mappings, countable and uncountable sets. Rela tion. Rela tion between sets A, B is a subset of the Cartesian product A  B. Cartesian product A  B is a set of all ordered pairs  a, b  , where a  A, b  B

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Mat h ematic al logi c

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  1. Mathematical logic Lesson 5 Relations, mappings, countable and uncountable sets Relation, function

  2. Relation Relation between sets A, B is a subset of the Cartesian product A  B. Cartesian productA  B is a set of all ordered pairsa, b, where aA, bB (Binary) relation R2on a set M is a subset ofM  M:R2 M  M n-ary relation Rnon a set M: Rn M ... M n times Relation, function

  3. Relation Mind: A couplea,b  b,a, but a set{a,b} = {b,a} a, aa, but {a,a} = {a} n-tuplesare ordered, particular elements of tuples do not have to be unique (can be repeated), unlike sets Notation: a,b  Ris written also in the prefix R(a,b)or infixwaya R b. For instance 1  3. Relation, function

  4. Relation - Example: Binary relationon the set of natural numbers N:< (strictly less than) {0,1,0,2,0,3,…,1,2,1,3, 1,4, …, 2,3,2,4,…,3,4,…,5,7,…,115,119, .…} Ternary relation on N: {0,0,0,1,0,1,1,1,0,…,2,0,2,2,1,1,2,2,0, …,3,0,3,3,1,2,3,2,1,3,3,0,…,115,110,5, .…}the set of triples of natural numbers such that the 3rd number equals the 1st minus the 2nd one Relation“adressof a person”: {Jan Novák, Praha 5, Bellušova 1831,Marie Duží, Praha 5, Bellušova 1827,...,} Relation, function

  5. Function (mapping) n-aryfunction F on a set M is a special“unique on the right-hand side” (n+1)-ary relation F  M ... M: (n+1) x abc ([F(a,b)  F(a,c)]  b=c) Partial F: to each n-tupleof elementsaM...M there existsat most one element bM. Notation F: M ... M  M, instead of F(a,b) we write F(a)=b. The set M ... M is called a domainof the function F, the set M is called a range. Relation, function

  6. Function (mapping) Example: Relation on N{1,1,1,2,1,2,2,2 ,1,…,4,2,2,…,9,3,3,…, 27,9,3, .…} Is a partial functiondividing without a remainder. The relationminus on N (see the previous slide) is a partial function on N: for instance the couple 2,4does not have an image in N. In order that the function minus were total, we’d have to extend the domain to integers. Relation, function

  7. Function (mapping) Functional symbols of FOL formulasare interpreted only by total functions: Total function F: A  BTo each element aA there is just one element bB such that F(a)=b: a b F(a)=b abc[(F(a)=b  F(a)=c)  b=c] Sometimes we introduce a special quantifier!With the meaning “there is just one”, written as: a !b F(a)=b Relation, function

  8. Function (mapping) Examples: Relation+{0,0,0, 1,0,1, 1,1,2, 0,1,1, …}is a(total binary) function on N. To each pair of numbers it assigns just one number, the sum of the former. Instead of 1,1,2 + we write 1+1=2. The relationis not a function: x y z [(x  y)  (x  z)  (y  z)] Relation{0,0, 1,1, 2,4, 3,9, 4,16, …}is a function on N, namely the total functionthe second power (x2) Relation, function

  9. Surjection, injection, bijection • A mappingf :A  B is called a surjection(mappingA ontoB), iff to each elementb  B there is an elementa  A such thatf(a)=b. • b [B(b)a (A(a) f(a)=b)]. • A mappingf :A  B is called an injection(one to onemappingA intoB), iff for all aA, bAsuch that a  b it holds thatf(a) f(b). • a b [(A(b)A(a)  (a  b)) (f(a)  f(b))]. • A mappingf :A  B is called a bijection(one to onemappingA ontoB), iff f is a surjection and injection. Relation, function

  10. Function (mapping) • Example: surjection injection bijection {1 2 3 4 5} {234 } {1 2 3 4 5} { 2 3 4 } {1 2 3 4 5} {1 2 3 4 5} If there is a bijection between the sets A, B, then we say that A and B have the same cardinality(number of elements). Relation, function

  11. Cardinality, countable sets • A set A that has the same cardinality as the set N of natural numbers is called acountable set. • Example: the set S of even numbers is countable. The bijection f of S into N is defined, e.g., byf(n) = 2n. Hence0  0, 1  2, 2  4, 3  6, 4  8, … One of the paradoxesof Cantor’s set theory: S  N (a proper subset) and yet the number of elements of the two sets is equal: Card(S) = Card(N)! Relation, function

  12. Cardinality, countablesets The set of rational numbers R is also countable. • Proof: in two steps. • Card(N)  Card(R), because each natural number is rational: N  R. • Now we construct amapping ofN ontoR (surjection N ontoR), by which we prove thatCard(R)  Card(N): 1 2 3 4 5 6 … 1/1 2/1 1/2 3/1 2/2 1/3 … But, in the table there are repeating rationals, hence the mapping is not one-to-one. However, no rational number is omitted, therefore it is a mapping of N onto R (surjection). Card(N) =Card(R). Relation, function

  13. Cardinality, uncountable sets • There are, however, uncountable sets: the least of them is theset of real numbersR • Even in the interval0,1there are more real numbers than the number of all natural numbers. However, in this interval there is the same number of reals than the number of all the realsR! • Cantor’s diagonal proof:If there were countably many real numbers in the interval 0,1, the numbers could be ordered into a sequence: the first one (1.), the second (2.), the third (3.),…, and each of these numbers would be of a form 0,i1i2i3…, where i1i2i3… is the decimal part of the number. • Rational numbers have a finite decimal part, irrational numbers have an infinite decimal part. • Let us add to each nthnumber in in the sequence i1i2i3… of decimals the number 1. We obtain a number which is not contained in the original sequence – see the next slide: Relation, function

  14. Cantor’s diagonal proof of uncountability of real numbers in the interval0,1. 1 2 3 4 5 6 7 1 i11 i12 i13 i14 i15 i16 i17 2 i21 i22 i23 i24 i25 i26 i27 3 i31 i32 i33 i34 i35 i36 i37 4 i41 i42 i43 i44 i45 i46 i47 5 i51 i52 i53 i54 i55 i56 i57 …. A new number that is not contained in the table: 0,i11+1 i22+1 i33+1 i44+1 i55+1 … Relation, function

  15. Propositional Logic again • Summary of the most important notions and methods. Relation, function

  16. Table of the truth functions Propositional Logic - summary

  17. Summary • Typical tasks: • Check whether an argument is valid • What is entailed by a given set of assumptions? • Add the missing assumptions so that the argument is valid • Is a given formula tautology, contradiction, satisfiable? • Find the models of a formula, find amodel of a set of formulas • Up to now we know the following methods: • Truth-table method • Equivalent transformations • An indirect semantic proof • The resolution method • Semantic tableau Propositional Logic

  18. Example. The proof of a tautology |= [(p  q)  (p  r)]  (q  r) Table:A

  19. Indirect proof of the tautology |= [(p  q)  (p  r)]  (q  r) The formulaA is a tautology, iff the negated formulaA is a contradiction: |= A iff A |= • Let us assume that the negated formula can be true. • Negation of implication: (A  B)  (A B) • (p  q)  (p  r) q r 1 1 1 0 1 0 q = 0, r = 0, hence p  0, p  0 0 0 0 0 therefore: p = 0, p = 0, i.e. 1 p = 1 contradiction • The negated formula does not have a model, it is a contradiction. Hence the formula A is a tautology. Propositional Logic

  20. The proof by equivalent transformations We need the laws: • (A  B)  (A  B)  ((A B)) • (A  B)  (A B) de Morgan • (A  B)  (A B) de Morgan • (A  B)  (A B) negation of implication • (A  (B  C))  ((A  B)  (A  C)) distributive law • (A  (B  C))  ((A  B)  (A  C)) distributive law • 1  A  1 1  tautology, • 1  A  A e.g. (p p) • 0  A  0 0 contradiction • 0  A  A e.g. (p p) propositional logic

  21. The proof by equivalent transformations |= [(p  q)  (p  r)]  (q  r) [(p  q)  (p  r)]  (q  r) [(p  q)  (p  r)]  (q  r)  • (p q)  (p r)  q  r  • [p  (p r)  q  r]  [q  (p r)  q  r]  • (p p q  r)  (p r q r)  (qp q r)  (q r  q  r)  1  1  1  1  1 – tautology • Note: We obtained a conjunctive normal form (CNF) Relation, function

  22. Proof of a tautology – resolution method |= [(p  q)  (p  r)]  (q  r) • Negated formula is transformed into a clausal form (CNF), the indirect proof: (p  q)  (p  r) q r  (p  q)  (p  r) q r 1. p  q 2. p  r 3. q 4. r 5. q  r resolution 1, 2 6. r resolution 3, 5 7. resolution 4, 6 – contradiction propositional logic

  23. Proof by a semantic tableau |= [(p  q)  (p  r)]  (q  r) Direct proof: we construct the CNF (‘’: branching, ‘’:comma – closed branches: ‘p p’) (p q)  (p  r)  q  r p, (p  r), q, r q, (p  r), q, r + p, p, q, r p, r, q, r + + propositional logic

  24. Indirect proof by a semantic tableau |= [(p  q)  (p  r)]  (q  r) Indirect proof: by the DNF of the negated formula (‘’: branching, ‘’:comma, - closed branches 0:‘p  p’) [(p q)  (p  r)]  q  r p, (p  r), q, r q, (p  r), q, r + p, p, q, r p, r, q, r + + propositional logic

  25. Proof of an argument |= [(p  q)  (p  r)]  (q  r) iff [(p  q)  (p  r)] |=(q  r) iff (p  q), (p  r) |=(q  r) p: The program goes right q: The system is in order r: It is necessary to call for a system engineer If the program goes right, the system is in order. If the program malfunctions, it is necessary to call for a system engineer ---------------------------------------------------------------------------- If the system is not in order, it is necessary to call for a system engineer. propositional logic

  26. Proof of an argument (p  q), (p  r) |=(q  r) Indirect proof: {(p  q), (p  r), (q  r)}– it cannot be a satisfiable set • p  q • p  r • q • r • q  r resolution 1, 2 • r resolution 3, 5 • resolution 4, 6, contradiction

  27. Proof of an argument (p  q), (p  r) |=(q  r) Direct proof: What is entailed by the assumptions? The resolution rule is truth preserving: p  q, p  r |-- q  r 1 1 1 In any valuation vit holds that if the assumptions are true, the resolvent is true as well Proof: a) p = 1  p = 0  q = 1  (q  r) = 1 b) p = 0  r = 1  (q  r) = 1 • p  q • p  r • q  r resolution 1, 2 – consequence: (q  r)  (q  r) QED propositional logic

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