3.9.2015Relation, function 1 Mathematical logic Lesson 5 Relations, mappings, countable and uncountable sets

19.04.2023 Relation, function 1

Mathematical logic

Lesson 5Relations, mappings, countable and uncountable sets

Relation, function 219.04.2023

Relation

Relation 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 aA, bB

(Binary) relation R2 on a set M is a subset ofM M: R2 M M

n-ary relation Rn on a set M: Rn M ... M

n times


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.


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 3rd number equals the 1st minus the 2nd one

Relation “adress of a person”: {Jan Novák, Praha 5, Bellušova 1831, Marie Duží, Praha 5, Bellušova 1827,...,}


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 bc ([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 bM.

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.


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 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.


Function (mapping)Functional symbols of FOL formulas are interpreted

only by total functions:

Total function F: A BTo each element aA there is just one element bB such that F(a)=b:

a b F(a)=b abc [(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


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 (x2)


Surjection, injection, bijection

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. b [B(b) a (A(a) f(a)=b)].

A mapping f : A B is called an injection (one to one mapping A into B), iff for all aA, bA such that a b it holds that f(a) f(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.


Function (mapping) Example:

surjection injection bijection

{1 2 3 4 5} {2 3 4 } {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).


Cardinality, countable sets A set A that has the same cardinality as the

set N of natural numbers is called a countable set.

Example: the set S of even numbers is countable. The bijection f of S into N is defined, e.g., by f(n) = 2n. 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)!


Cardinality, countable setsThe set of rational numbers R is also

countable.

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).

Card(N) = Card(R).

1/1 1/2 1/3 1/4 1/5 1/6 …

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 …

… … … … … … …


Cardinality, uncountable sets

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,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 nth number 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:


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 …


Propositional Logic again

Summary of the most important notions and methods.

Propositional Logic - summary 1619.04.2023

Table of the truth functions

A B A A B A B A B A B

1 1 0 1 1 1 1

1 0 0 1 0 0 0

0 1 1 1 0 1 0

0 0 1 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.

Propositional Logic 1719.04.2023

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 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

Example. The proof of a tautology |= [(p q) (p r)] (q r)

Table: A

p q r (p q) (p r) A (q r) A (q r)

1 1 1 1 1 1 1 1

1 1 0 1 1 1 1 1

1 0 1 0 1 0 1 1

1 0 0 0 1 0 0 1

0 1 1 1 1 1 1 1

0 1 0 1 0 0 1 1

0 0 1 1 1 1 1 1

0 0 0 1 0 0 0 1

Propositional Logic 1919.04.2023

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 |= 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 = 1contradiction

The negated formula does not have a model, it is a contradiction. Hence the formula A is a tautology.

propositional logic 2019.04.2023

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)


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)


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 q2. p r3. q4. r5. q r resolution 1, 26. r resolution 3, 57. resolution 4, 6 – contradiction


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

+ +


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

+ +


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 rightq: The system is in orderr: 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.

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

1. p q2. p r3. q4. r5. q r resolution 1, 26. r resolution 3, 57. resolution 4, 6, contradiction


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 wellProof:a) p = 1 p = 0 q = 1 (q r) = 1b) p = 0 r = 1 (q r) = 11. p q2. p r3. q r resolution 1, 2 – consequence:(q r) (q r) QED

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3.9.2015Relation, function 1 Mathematical logic Lesson 5 Relations, mappings, countable and uncountable sets