Mathematical logic Lesson 5 Relations mappings countable and
Mathematical logic Lesson 5 Relations, mappings, countable and uncountable sets 6/19/2021 Relation, function 1
Relation between sets A, B is a subset of the Cartesian product A B is a set of all ordered pairs a, b , where a A, b B (Binary) relation R 2 on a set M is a subset of M M: R 2 M M n-ary relation Rn on a set M: Rn M . . . M n times 6/19/2021 Relation, function 2
Relation Mind: A couple a, b b, a , but a set {a, b} = {b, a} a, a a , but {a, a} = {a} n-tuples are ordered, particular elements of tuples do not have to be unique (can be repeated), unlike sets Notation: a, b R is written also in the prefix R(a, b) or infix way a R b. For instance 1 3. 6/19/2021 Relation, function 3
Relation - Example: Binary relation on 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 3 rd number equals the 1 st minus the 2 nd one Relation “adress of a person”: { Jan Novák, Praha 5, Bellušova 1831 , Marie Duží, Praha 5, Bellušova 1827 , . . . , } 6/19/2021 Relation, function 4
Function (mapping) n-ary function 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-tuple of elements a M. . . M there exists at 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 domain of the function F, the set M is called a range. 6/19/2021 Relation, function 5
Function (mapping) Example: Relation on N { 1, 1 , 2, 1 , 2 , 2, 2 , 1 , …, 4, 2 , …, 9, 3 , …, 27, 9 , 3 , . …} Is a partial function dividing without a remainder. The relation minus 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. 6/19/2021 Relation, function 6
Function (mapping) Functional symbols of FOL formulas are interpreted only by total functions: Total function F: A B To 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 6/19/2021 Relation, function 7
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 function the second power (x 2) 6/19/2021 Relation, function 8
Surjection, injection, bijection n A mapping f : A B is called a surjection (mapping A onto B), iff to each element b B there is an element a A such that f(a)=b. ¨ n A mapping f : A B is called an injection (one to one mapping A into B), iff for all a A, b A such that a b it holds that f(a) f(b). ¨ n b [B(b) a (A(a) f(a)=b)]. a b [(A(b) A(a) (a b)) (f(a) f(b))]. A mapping f : A B is called a bijection (one to one mapping A onto B), iff f is a surjection and injection. 6/19/2021 Relation, function 9
Function (mapping) n Example: surjection {1 2 3 4 5} injection {2 3 4 } bijection {1 2 3 4 5} {234} {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). 6/19/2021 Relation, function 10
Cardinality, countable sets A set A that has the same cardinality as the set N of natural numbers is called a countable set. n Example: the set S of even numbers is countable. The bijection f of S into N is defined, e. g. , by f(n) = 2 n. Hence 0 0, 1 2, 2 4, 3 6, 4 8, … One of the paradoxes of 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)! n 6/19/2021 Relation, function 11
Cardinality, countable sets The set of rational numbers R is also countable. 1/1 1/2 1/3 1/4 1/5 1/6 … Proof: in two steps. a) Card(N) Card(R), because each natural number is rational: N R. b) Now we construct a mapping of N onto R (surjection N onto R), by which we prove that Card(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). n 2/1 2/2 2/3 2/4 2/5 2/6 … 3/1 3/2 3/3 3/4 3/5 3/6 … 4/1 4/2 4/3 4/4 4/5 4/6 … 5/1 5/2 5/3 5/4 5/5 5/6 … 6/1 6/2 6/3 6/4 6/5 6/6 … … … … Card(N) = Card(R). 6/19/2021 Relation, function 12
Cardinality, uncountable sets n n n There are, however, uncountable sets: the least of them is the set of real numbers R 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 reals R! 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, i 1 i 2 i 3…, where i 1 i 2 i 3… 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 nth number in in the sequence i 1 i 2 i 3… of decimals the number 1. We obtain a number which is not contained in the original sequence – see the next slide: 6/19/2021 Relation, function 13
Cantor’s diagonal proof of uncountability of real numbers in the interval 0, 1. 1 2 i 12 i 22 i 32 i 42 i 52 3 i 13 i 23 i 33 i 43 i 53 4 i 14 i 24 i 34 i 44 i 54 5 i 15 i 25 i 35 i 45 i 55 6 i 16 i 26 i 36 i 46 i 56 7 i 17 i 27 i 37 i 47 i 57 1 i 11 2 i 21 3 i 31 4 i 41 5 i 51 …. A new number that is not contained in the table: 0, i 11+1 i 22+1 i 33+1 i 44+1 i 55+1 … 6/19/2021 Relation, function 14
Propositional Logic again n Summary of the most important notions and methods. 6/19/2021 Relation, function 15
Table of the truth functions A 1 1 0 0 B 1 0 A 0 0 1 1 A B A B 1 1 1 0 0 0 1 1 Be careful with implication, p q. It is false only in one case: p = 1, q = 0. It is something like a promise: “If you behave well you will get a Christmas gift” (p q). “I have been a good boy but there is no Christmas gift”. (p q) Has the promise been fulfilled? If he were not a good boy (p = 0), then the promise would not obligate to anything. 6/19/2021 Propositional Logic - summary 16
Summary n Typical tasks: ¨ ¨ ¨ n 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 a model 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 6/19/2021 Propositional Logic 17
Example. The proof of a tautology |= [(p q) ( p r)] ( q r) A Table: p q r (p q) ( p r) A ( q r) A ( q r) 1 1 1 1 1 0 1 0 1 1 1 0 0 0 1 0 1 1 1 1 0 1 0 0 1 1 1 1 0 0 0 1
Indirect proof of the tautology |= [(p q) ( p r)] ( q r) The formula A is a tautology, iff the negated formula A is a contradiction: |= A iff A |= n Let us assume that the negated formula can be true. n Negation of implication: (A B) n (p q) ( p r) q r 1 1 10 10 q = 0, r = 0, hence p 0, p 0 0 0 therefore: p = 0, i. e. 1 p=1 contradiction n The negated formula does not have a model, it is a contradiction. Hence the formula A is a tautology. 6/19/2021 Propositional Logic 19
The proof by equivalent transformations We need the laws: n (A B) ( (A B)) n (A B) ( A B) de Morgan n (A B) negation of implication n (A (B C)) ((A B) (A C)) distributive law n 1 A 1 n 1 A A n 0 A 0 n 0 A A 6/19/2021 1 tautology, e. g. (p p) 0 contradiction e. g. (p p) propositional logic 20
The proof by equivalent transformations |= [(p q) ( p r)] ( q r) Û (p q) ( p r) q r Û [p ( p r) q r] [ q ( p r) q r] Û (p p q r) (p r q r) ( q p q r) ( q r) 1 1 1 – tautology n Note: We obtained a conjunctive normal form (CNF) 6/19/2021 Relation, function 21
Proof of a tautology – resolution method |= [(p q) ( p r)] ( q r) n 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. 2. 3. 4. 5. 6. 7. p q p r q r resolution 1, 2 r resolution 3, 5 resolution 4, 6 – contradiction 6/19/2021 propositional logic 22
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, q, r + 6/19/2021 p, r, q, r + propositional logic 23
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 + 6/19/2021 p, r, q, r + propositional logic 24
Proof of an argument |= [(p q) ( p r)] ( q r) [(p q) ( p r)] |= ( q r) (p q), ( p r) |= ( q r) iff 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. 6/19/2021 propositional logic 25
Proof of an argument (p q), ( p r) |= ( q r) Indirect proof: {(p q), ( p r), ( q r)} – it cannot be a satisfiable set 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 26
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 v it 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 1. p q 2. p r 3. q r resolution 1, 2 – consequence: (q r) ( q r) QED 6/19/2021 propositional logic 27
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