1. Lesson 5 Relations, mappings, countable and uncountable sets
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 product A  B is a set of all ordered pairs a, b, where aA, bB
(Binary) relation R2 on a set M is a subset of M  M: R2  M  M
n-ary relation Rn on 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-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, function

4. 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,...,}
Relation, function

5. 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.
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 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.
Relation, function

7. 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
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 function the second power (x2)
Relation, function

9. 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 aA, bA 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.
Relation, function

10. Function (mapping) Example: surjection injection bijection
{ } {2 3 4 } { }
{ } { } { }
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 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)!
Relation, function

12. Cardinality, countable sets
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 a mapping of N onto R (surjection N onto R), by which we prove that Card(R)  Card(N): …
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
Relation, function

13. 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:
Relation, function

14. A new number that is not contained in the table:
Cantor’s diagonal proof of uncountability of real numbers in the interval 0,1.
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 functions1
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 - summary

17. Summary Typical tasks: Up to now we know the following methods:
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
Propositional Logic

18. 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

19. 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
q = 0, r = 0, hence p  0, p  0
therefore: p = 0, p = 0, i.e.
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   tautology,
1  A  A e.g. (p  p)
0  A   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: 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
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
p  q
p  r
q  r resolution 1, 2 – consequence:
(q  r)  (q  r) QED
propositional logic