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

2. Relation, function 2 3.9.2015 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 R 2 on a set M is a subset of M  M: R 2  M  M n-ary relation R n on a set M: R n  M ...  M n times

3. Relation, function 3 3.9.2015 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.

4. Relation, function 4 3.9.2015 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 ,...,}

5. Relation, function 5 3.9.2015 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. Relation, function 6 3.9.2015 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.

7. Relation, function 7 3.9.2015 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

8. Relation, function 8 3.9.2015 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 )

9. Relation, function 9 3.9.2015 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.

10. Relation, function 10 3.9.2015 Function (mapping) Example: surjectioninjectionbijection {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).

11. Relation, function 11 3.9.2015 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)!

12. Relation, function 12 3.9.2015 Cardinality, countable sets The 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). 1/11/21/31/41/51/6… 2/12/22/32/42/52/6… 3/13/23/33/43/53/6… 4/14/24/34/44/54/6… 5/15/25/35/45/55/6… 6/16/26/36/46/56/6… …………………

13. Relation, function 13 3.9.2015 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,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 n th number i n 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:

14. Relation, function 14 3.9.2015 Cantor’s diagonal proof of uncountability of real numbers in the interval  0,1 . 1234567 1 i 11 i 12 i 13 i 14 i 15 i 16 i 17 2 i 21 i 22 i 23 i 24 i 25 i 26 i 27 3 i 31 i 32 i 33 i 34 i 35 i 36 i 37 4 i 41 i 42 i 43 i 44 i 45 i 46 i 47 5 i 51 i 52 i 53 i 54 i 55 i 56 i 57 …. 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 …

15. Relation, function 15 3.9.2015 Propositional Logic again Summary of the most important notions and methods.

16. Propositional Logic - summary 16 3.9.2015 Table of the truth functions A B A AA  BA  BA  BA  B 1 1 01111 1 0 01000 0 1 11010 0 0 10011 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.

17. Propositional Logic 17 3.9.2015 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

18. Example. The proof of a tautology |= [(p  q)  (  p  r)]  (  q  r) Table: A pqr (p  q)(  p  r) A (  q  r)A  (  q  r) 11111111 11011111 10101011 10001001 01111111 01010011 00111111 00010001

19. Propositional Logic 19 3.9.2015 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 = 1 contradiction The negated formula does not have a model, it is a contradiction. Hence the formula A is a tautology.

20. propositional logic 20 3.9.2015 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  11  tautology, 1  A  Ae.g. (p   p) 0  A  00  contradiction 0  A  Ae.g. (p   p)

21. Relation, function 21 3.9.2015 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)

22. propositional logic 22 3.9.2015 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  rresolution 1, 2 6.rresolution 3, 5 7.resolution 4, 6 – contradiction

23. propositional logic 23 3.9.2015 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+

24. propositional logic 24 3.9.2015 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 ++

25. propositional logic 25 3.9.2015 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.

26. 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  q 2. p  r 3.  q 4.  r 5. q  rresolution 1, 2 6. rresolution 3, 5 7. resolution 4, 6, contradiction

27. propositional logic 27 3.9.2015 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  rresolution 1, 2 – consequence: (q  r)  (  q  r)QED