Mathematical logic

1
Mathematical logic

• Lesson 5
• Relations, mappings,
• countable and uncountable sets

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 ofM
  ? M R2 ? M ? M
• n-ary relation Rn on a set M Rn ? M ?...? M
• n times

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.

4
Relation - Example

• Binary relation on the set of natural numbers N
  lt (strictly less than) ?0,1?,?0,2?,?0,3?,,?1,2?
  ,?1,3?, ?1,4?, , ?2,3?,?2,4?,,?3,4?,,?5,7?,,?1
  15,119?, .
• 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?,...,

5
Function (mapping)

• n-ary function F on a set M is a special unique
  on the right-hand side (n1)-ary relation F ? M
  ?...? M
• (n1) x
• ?a ?b?c (F(a,b) ? F(a,c) ? bc)
• 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
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, wed have to extend the domain to integers.

7
Function (mapping)

• Functional symbols of FOL formulas are
  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) ? bc
• Sometimes we introduce a special quantifier ?!
  With the meaning there is just one, written
  as
• ?a ?!b F(a)b

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 112.
• 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)

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.

10
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).

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 Cantors set theory S ?
  N (a proper subset) and yet the number of
  elements of the two sets is equal Card(S)
  Card(N)!

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 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

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!
• Cantors 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

14
Cantors 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,i111 i221 i331 i441 i551

15
Propositional Logic again

• Summary of the most important notions and methods.

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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.
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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

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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
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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
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)

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)

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

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

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

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.

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 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