Computing Fundamentals 1 Lecture 8 Functions

1
Computing Fundamentals 1 Lecture 8Functions

• Lecturer Patrick Browne
• http//www.comp.dit.ie/pbrowne/
• Room K308
• Based on Chapter 14.
• A Logical approach to Discrete Math
• By David Gries and Fred B. Schneider

2

• Given two countries Aland and Bland. We can
  travel from Aland to Bland using AtoB airways.
  Cities in Aland are called ai and cities in Bland
  are called bi .

b4
a1 a2 a3
AtoB
b1 b2 b3
Aland
Bland
3

• From any given city ai in Aland AtoB airways
  provide one and only one flight to one city bi in
  Bland. In other words, you can travel from every
  city in Aland but cannot travel to more than one
  city in Bland from that Aland city. We can
  consider the flights offered by AtoB as function
  called FAtoB.

b4
a1 a2 a3
FAtoB
b1 b2 b3
Aland
Bland
4

• There are some cities in Bland, such as b4, that
  are not served by AtoB airways, therefore they
  cannot be reached from Aland. Obviously, if you
  cannot get to b4 then you cannot come back to
  Aland from b4 . (no return tickets)

b4
a1 a2 a3
AtoB
b1 b2 b3
Aland
Bland
5

• If for every city that AtoB airways flies into
  in Bland they supply a return ticket, then AtoB
  supply a pair of functions, FAtoB and FBtoA. We
  call FAtoB an injective function because it has a
  left inverse (a return ticket).

b4
a1 a2 a3
FAtoB
b1 b2 b3
Aland
Bland
6

• If AtoB airways flies into every city in Bland,
  still using the original rule that only one
  flight can leave an Aland city, then there may be
  more than one way back to Aland. For example,
  from b2 you can fly to a1 or a3. Then FAtoB is
  called a surjective function. It has a right
  inverse (but you may not get back to where you
  started).

a1 a2 a3
FAtoB
b1 b2
Aland
Bland
7

• If a function is both injective and surjective it
  is then called bijective, it has both a left and
  right inverse.

a1 a2 a3
FAtoB
b1 b2 b3
Aland
Bland
8
http//en.wikipedia.org/wiki/Inverse_functionLeft
_and_right_inverses
9
Functions

• We apply function to argument (function
  application)
• Function definition g.x 3 ? x 6
• g(5)
• Gives the value of 3?56
• To reduce brackets we can write
  function.argument.
• We evaluate this function
• g.5
• lt Apply functiongt
• 3 ? 5 6
• lt Arithmeticgt
• 21

10
Functions

• Functions can be considered as a restricted form
  of relation. This is useful because the
  terminology and theory of relations carries over
  to function.
• In programming languages like C or CafeOBJ a
  function can have a signature, which includes its
  name, type of argument and the type of the
  expected return value.

11
Fibonacci Function in C and CafeOBJ
Signature of a function consists of a name,
argument(s) type, and return type Function Name
Argument type p is a
predecessor function
mod FIBO-NAT pr(NAT) op fib Nat -gt Nat
var N Nat eq fib(0) 0 . eq fib(1) 1 .
ceq fib(N) fib(p(N)) fib(p(p(N)))
if N gt 1 .
int fib(int n) if (n lt 1) return n else
return fib(n-1) fib(n-2)
Argument variable
12
Fibonacci Function in C and CafeOBJ
Signature of a function consists of a name,
argument(s) type, and return type.
Return type
mod FIBO-NAT pr(NAT) op fib Nat -gt Nat
var N Nat eq fib(0) 0 . eq fib(1) 1 .
ceq fib(N) fib(p(N)) fib(p(p(N)))
if N gt 1 .
int fib(int n) if (n lt 1) return n else
return fib(n-1) fib(n-2)
Argument constraints
13
Functions

• A function f is a rule for computing a value v
  from another value w, so that the application
  f(w) or f.w denotes a value vf.w v. The
  fundamental property of function application,
  stated in terms of inference rule Leibniz is

• This property allows us to conclude theorems like
  f(bb) f(2?b) and f.b f.b 2?f.b.

14
Functions and programming

• In programming functions can have side effects
  i.e. they can change a parameter or a global
  variable. For example, if b is changed by the
  function f then
• f.b f.b 2 ? f.b
• no longer holds. This makes it difficult to
  reason or prove properties about a program. By
  prohibiting side-effects we can use mathematical
  laws for reasoning about programs involving
  function application.

15
Functions as relations

• While we can consider functions as a rule we can
  also think of functions as a binary relation B ?
  C, that contains all pairs ltb,cgt such that f.bc.
  A relation can have distinct values c and c that
  satisfy bfc and bfc, but a function cannot.

c
Allowed for relations but not allowed for
functions
b
c
16
Functions Textual Substitution

• Function application can be defined as textual
  substitution. If
• g.zE
• Defines a function g then function application
  g.X for any argument X is defined by
• g.X Ez X

17
Function Def. Types

• A binary relation f on B ? C, is called a
  function iff it is determinate.
• Determinate
• (?b,c,c b f c ? b f c cc)
• A function f on B ? C is total if
• Total B Dom.f
• Otherwise it is partial. We write fB?C for the
  type of f if f is total and fB?C if f is
  partial.

18
Functions

• This close correspondence between function
  application and textual substitution suggests
  that Leibniz (1.5) links equality and function
  application. So, we can reformulate Leibniz for
  functions.

19
Functions Types

• In Computer Science many types are needed.
• Simple types like integers and natural numbers
• Complex types like sets, lists, sequences.
• Each function has a type which describes the
  types of its parameters and the type of its
  result
• (8.1) f t1 ? t1 ...? t1? r

20
Functions their types

• plus????? (plus(1,3)or 13)
• not ??? (not(true) or ?true)
• less????? (less(1,3)or 1lt3)

21
Functions Types
22
Functions Types

• Certain restrictions are need to insure
  expression are type correct.
• During textual substitution Ex F, x and F
  must have the same type.
• Equality bc is defined only if b and c have the
  same type. Treating equality as an infix
  function
• __ t?t? ?
• For any type t.

23
14.41 Theorem

• Total function fB?C is surjective (or onto ) if
  Ran.f C.
• Total function fB?C is injective (or one-to-one)
  if
• (?b,bB,cC bfc ? bfc ? bb)
• A function is bijective if it is injective and
  surjective.

24
14.41 Theorem

• Total function f is injective,one-to-one if
• (?b,bB,cC bfc ? bfc ? bb)
• Total function fB?C is surjective, onto if
• Ran.f C.
• A function is bijective if it is injective and
  surjective.

25

• An Injective function can
• f A -gtB fA -gtB f-1 ? B -gtA f (have
  inverse)
• f A-gtB fA -gtB dom f A f (be
  total)

Target
Source
f
a1 a2 a3
b1 b4 b2 b3
A
B
dom f
ran f
26
Injective function in CafeOBJ()

• module INJ
• A B
• op f_ A -gt B
• op g_ B -gt A
• var a A
• vars a b' B
• eq linv g f a a .
• ceq inj b b' if g b g b' .

27
Injective function in CafeOBJ()

• eq linv g f A A .
• linv represents an axiom for a left inverse
  taking an A to a B and back to the same A. The
  linv equation says that g is a left inverse of
  f (i.e. g(f(a)) a).
• ceq inj B B' if g B g B' .
• The conditional equation inj represents the
  injective property, that two Bs are the same if
  they map to the same A. The inj equation
  expresses the injective (or one-to-one) property.

28

• CafeOBJ Injective Function left inverse
• mod INJ A B
• op f_ A -gt B op g_ B -gt A
• var a A
• vars b0 b1 B
• eq linv g f a a .
• ceq inj b0 b1 if g b0 g b1 .

Not in range of f.
b4
a1 a2 a3
f
b1 b2 b3
A
B
dom f
ran f
29

• Surjective Function
• Total function fA?B is surjective, onto if
• Ran.f B.

a1 a2 a3
f
b1 b2
A
B
dom f
ran f
30
14.42 Theorem

• Let fB?C be a total function, and let f-1 is
  its relational inverse. If f is not injective
  (one-to-one) then f-1 is not a determinate
  function.

?
?
?
B
C
?
C
B
?
?
Function f not injective
The inverse (f-1) is not determinate
31
14.42 Theorem

• Let fB?C be a total function, and let f-1 be its
  relational inverse. Then f-1 is a (i.e.
  determinate) function iff f is injective
  (one-to-one). And, f-1 is total if f is
  surjective (onto).

?
?
?
?
B
B
C
C
?
?
Inverse not surjective (onto)
Function not total
32
14.42 Theorem

• Let fB?C be a total function, and let f-1 be its
  relational inverse. Then f-1 is total iff f is
  surjective (onto).

?
?
?
?
B
B
C
C
?
?
Inverse not surjective (onto)
Function not total
33
Total Partial functions

• Dealing with partial functions can be difficult.
  Whats the value of (f.bf.b) if b?Dom.f ?
• The choice of value must be such that the rules
  of manipulations that we use in the propositional
  and predicate calculus hold even in the presence
  of undefined values, and this is not easy to
  achieve. However, for partial function fB?C one
  can always restrict attention to its total
  counterpart, fDom.f? C

34
Functions

• Binary relation lt is not a function because 1
  lt 2 and 1 lt 3 both hold.
• Identity relation iB over B is a total function,
  iBB?B ibb for all b in B.
• Total function f??? is defined by f(n)n1 is
  the relation lt0,1gt,lt1,2gt,.
• Partial function f??? is defined by f(n) 1/n
  is the relation lt1,1/1gt,lt2,1/2gt,lt3,1/3gt. It is
  partial because f.0 is not defined.
• Note ? is a natural number, ? is a rational
  number.

35
Functions

• ? is an integer, ? is a positive integer, ?- is
  a negative integer.
• Function f??? is defined by f(b)1/b is total,
  since f.b is defined for all elements of ?.
  However, g??? is defined by g.b 1/b is partial
  because g.0 is not defined.

36
Functions

• The partial function f takes each lower case
  character to the next character can be defined by
  a finite number of pairs lta,bgt,ltb,cgt,..,lt
  y,zgt
• It is partial because there is no component whose
  first component is z.

37
Functions

• When partial and total functions are viewed as
  binary relations, functions can inherit
  operations and properties of binary relations.
• Two functions are equal when their sets of pairs
  are equal.
• Similar to relations, we can have the product and
  powers (or composition see later) of functions.

38
Inverses of Total Functions

• Every relation has an inverse relation.
• However, the inverse of a function does not have
  to be a function. For example
• f??? defined as f(b)b2.
• f(-2)4 and f(2)4
• f-1(4)?2, two values.

39
Inverse of Total Functions

• Partial functions
• Total functions
• Injective or on-to-one
• Surjective or onto
• Bijective has inverse

40
Functions Products

• We can have the product of two relations.
• We now look at the product (f ? g) of two total
  functions.
• (f ? g).b d
• lt viewing as relation gt
• b(f ? g)b
• lt product of relation 14.20gt
• (? bfc ? cgd)
• lt relation as function application gt
• (? f.bc ? g.cd)
• ltTrading 9.19gt
• (? c f.b g.cd)
• lt one point rule 8.14gt
• g(f.b) d

41
Functions Products

• (f ? g).b d
• lt viewing as relation gt
• b(f ? g)b
• lt product of relation 14.20gt
• (? bfc ? cgd)
• lt relation as function application gt
• (? f.bc ? g.cd)
• ltTrading 9.19gt
• (? c f.b g.cd)
• lt one point rule 8.14gt
• g(f.b) d
• Hence (f ? g).b g(f.b). This illustrates the
  difference between relational notation and
  functional notation.

42
Functions Products

• Hence (f?g).b g(f.b). This illustrates the
  difference between relational notation and
  functional notation, particularly for products.
  Functions possess an asymmetry between input and
  output. Expression f(arg) stands for the value of
  the function for the argument. On the other hand
  a relational expression r(a,b) is a statement
  that may be true or false. The product operation
  for relations is legal when the range of the
  first relation is the same as the domain of the
  second relation.

43
Functions Products

• We would prefer f(g.b) instead of g(f.b). So we
  have the new symbol ? for composition of
  functions, defined as
• f ? g ? g ? f

44
Composition of Functions1
g o f, the composition of f and g
45
Composition of Functions1

• The functions fX?Y and gY?Z can be composed by
  first applying f to an argument x and then
  applying g to the result. Thus one obtains a
  function
• gof X ? Z
• defined by (gof)(x)  g(f(x)) for all x in X.
• The notation gof is read as "g circle f" or "g
  composed with f.
• The composition of functions is always
  associative. That is, if f, g, and h are three
  functions with suitably chosen domains and
  codomains, then f o (g o h)  (f o g) o h.

46
A function in C

• int addTwo(int arg) Example of
  usage
• return (arg2) y addTwo(7)

addTwo(arg1)
arg1 arg2
return1 return2
addTwo(arg2)
Return valueint
Argumentint
47

• A function is a relation
• A function is a special case of a relation in
  which there is at most one value in the range for
  each value in the domain. A function with a
  finite domain is also known as a mapping. Note in
  diagram no diverging lines from left to right.

f
x5
x1 x2 x4 x6
y1 y2
X
Y
x3
Dom.f
Ran.f
48

• A function is a relation
• The function fX?Y the function f, from X to
  Y
• Is equivalent to relation XfY with the
  restriction that for each x in the domain of f, f
  relates x to at most one y

f
x5
x1 x2 x4 x6
y1 y2
X
Y
x3
Dom.f
Ran.f
49
Functions

• The relation between persons and their identity
  numbers
• identityNo ? lt Person, ?gt or
• Person identityNo ?
• is a function if there is a rule that a person
  may only have one identity number . It could be
  defined
• identityNo Person ? ?
• Perhaps there should also be a rule that only one
  identity number may be associated with any one
  person, but that is not indicated here.

50
Functions

• A functions source and target can be the same
  domain or type
• hasMother Person ? Person
• any person can have only one mother and several
  people could have the same mother.

51
Function Application

• All the concepts which pertain to relations apply
  to functions.
• In addition a function may be applied.
• Since there will be at most only one value in the
  range for a given x it is possible to designate
  that value directly.
• The value of f applied to x is the value in the
  range of function of the function f corresponding
  to the value of x in its domain.

52
Function Application

• The application is undefined if the value of x is
  not in the domain of f.
• It is important to check that the value of x is
  in the domain of the function f before
  attempting to apply f to x.
• The application of the function to the value x
  (the argument) is written f.x or f(x)

53
Function Application

• We can check the applicability of f by writing
  
• x ? dom f
• f.x y
• These predicates must be true if f(x)y is a
  function.

54
Partial and Total Functions

• If the identyNo function is a partial function
  there may be values in the source that are not in
  the domain (some people may have no identity
  number).
• A total function is one where there is a is a
  value for every possible value of x, so f.x is
  always defined. It is written
• f X ?Y
• Because
• Dom f X (or source)
• function application can always be used with a
  total function.

55
Total Functions

• The function age, from Person to natural number,
  is total since every person has an age
• age Person ? ?
• The function hasMother, is total since every
  person has one mother (including a mother!)
• hasMother Person ? Person

56
Injective Function

• An injective function may be partial or total.
• The functions identityNo and MonogamousMarriage
  are injective.

57
Isomorphism

• Given two functions
• fA-gtB, gB-gtA
• f is isomorphism or invertible map iif gfIA and
  fgIB .
• Where IA, IB are the identity functions of A and
  B respectively.
• A left inverse of f is a function g such that
• g(f(x)) x.
• or using alternative notation
• g f x x

58
Isomorphism

• Let fA-gtA and gB-gtB be two functions on A and B
  respectively. If there is a mapping t that
  converts all x in A to some ut(x) in B such that
  g(t(x))t(f(x)), then the two functions f and g
  are homomorphic and t is a homomorphism
• If also if t is bijective, then the two
  functions are isomorphic and t is an isomorphism
  with respect to f and g.

59
Isomorphism
x t u

Arabic f
g Roman
y t z

60

• An Injective functions1.

Two injective functions
A non-injective
function.
61
Injective Function f

• Using the INJ module and the CafeOBJ apply
  command we will prove that an injective function
  with a right inverse is an isomorphism.
• For fA-gtB, gB-gtA is isomorphism or invertible
  map iif gfIA and fgIB
• A function f has at least one left inverse if it
  is an injection.
• A left inverse is a function g such that
• g f a a
• or
• g(f(a)) a
• See lab 6 for details

62
Injective Function f

• module INJ -- module name
• A B -- two sorts
• op f_ A -gt B -- a function from A to B
• op g_ B -gt A -- a function from B to A
• var a A -- variables
• vars b b' B
• -- equation for left inverse equation of f
• eq linv g f a a .
• -- conditional equation for injective property
• ceq inj b b' if g b g b' .
• The Linv equation says that g(f(x)) x, i.e. g
  is a left inverse of f.
• The inj equation expresses the injective (or
  one-to-one) property.
• A function f has at least one "left inverse" if
  and only if it is an injection.

63
Injective Function f

• open INJ .
• op b gt B . -- make a constant b
• -- make a right inverse as the term to which the
  equation(s) will be applied to.
• start f g b .
• -- substitute B f g b (our term) and B' b
  in inj
• apply .inj with B' b at term .
• -- normal evaluation will execute linv
• apply red at term .
• result b B

64
Injective Function f

• In CafeOBJ equations are written declaratively
  and interpreted operationally as rewrite rules,
  which replace substitution instances of the LHS
  by the corresponding substitutions instances of
  the RHS. The operational semantics normally
  require that the least sort of the LHS is greater
  that or equal to the least sort of the RHS. The
  Linv equation says that g(f(x)) x (or g is a
  left inverse of f). The inj equation expresses
  the injective (or one-to-one) property.

65
Right Inverse

• module GROUP
• G
• op 0 -gt G
• op __ G G -gt G assoc
• op -_ G -gt G
• var X G
• eq EQ1 0 X X .
• eq EQ2 (-X) X 0 .

66
Right Inverse manual Proof

• We want to prove that X is a right inverse of .
• A standard proof goes as follows
• X ( X)
• 0 X ( X)
• ( ( X)) ( X) X ( X)
• ( ( X)) 0 ( X)
• ( ( X)) ( X)
• 0

67
Right inverse CafeOBJ proof

• open GROUP
• op a gt G . -- a constant
• start a ( a) . the term
• apply .1 at (1) .
• apply .2 with X a at 1 .
• apply reduce at term .

68

• Injective composition.

69

• An Surjective functions1.

Two surjective functions
A non-surjective
function
70

• An Surjective Composition1.

• Surjective composition the first function need
  not be surjective

71

• A bijection 1.

A bijective function
Composition gof of two functions is bijective, we
can only say that f is injective and g is
surjective.
72
Some examples, range on the left domain on the
right
A function is called a bijection , if it is onto
and one-to-one. A bijective function has an
inverse
Function Application
A function f from X to Y is a relation from X to
Y having the properties 1. The domain of f is
in X. 2. If (x,y),(x,y)?f, then yy (no fan out
from domain). A total function from X to Y is
denoted f X? Y.
73
A Function

• Given
• Side left ? right
• the function
• driveON Country ? Side
• Representing the side of the road vehicles drive
  on is a surjection, because the range includes
  all the values of the type Side. It would usually
  be considered total because driveOn should be
  defined in all countries.

74
What is a Hash Function?

• A hash function takes a data item to be stored or
  retrieved (say to/from disk) and computes the
  first choice for a storage location for the item.
  For example assume StudentNumber is the key
  field of a student record, and one record is
  stored on one disk block. To store or retrieve
  the record for student number n in a choice of
  11 disk locations numbered 0 to 10 we might use
  the hash function.
• h(n) n mod 11

75
Describe the steps needed to store a record for
student number 257 using in locations disk 0-10
using h(n) n mod 11
558
32
132
5
15
102
0 1 2 3 4 5 6
7 8 9 10
Steps to store 257 using h(n)n mod 11 into
locations 0-10 1. Calculate h(257) giving 4. 2.
Note location is already occupied, a collision
has occurred. 3. Search for next free space ,
with 0 assumed to follow 10. 4. If no free space
then array is full, otherwise store 257 in next
free space
558
32
132
5
15
102
257
0 1 2 3 4 5 6
7 8 9 10
76
Exercise 1

• Only one person can book a room. A system records
  the booking of hotel rooms on one night. Given
  the domain of discourse consists of
• Rooms the set of all rooms in hotel
• Person the set of all possible persons
• The state of the hotels bookings can be
  represented by the following
• bookedTo Room ? Person

77
Exercise 1 Solution

• (a) Explain why bookedTo is a function.
• bookedTo is a function since it maps rooms to
  persons and for any given room(domain) at most
  one person (range) can book it. A person can book
  any number of rooms.
• b) Explain why bookedTo is partial.
• The function is partial since not all rooms have
  been booked.

78
Example()

• Which of the following sets P, Q, and R
  represents a partial, surjective, injective or
  bijective function from X1,2,3,4 to
  Ya,b,c,d? In each case explain your reasoning.
  
• P lt1,agt,lt2,agt,lt3,cgt,lt4,bgt
• Q lt1,cgt,lt2,agt,lt3,bgt,lt4,cgt,lt2,dgt
• R lt1,cgt ,lt2,dgt,lt3,agt,lt4,bgt

79
Example()

• Which of the following sets P, Q, and R
  represents a partial, surjective, injective or
  bijective function from X1,2,3,4 to
  Ya,b,c,d? In each case explain your reasoning.
  
• P lt1,agt,lt2,agt,lt3,cgt,lt4,bgt
• P is a total function from X to Y.
• Domain X, rangea,b,c. The function neither
  injective (both 1 and 2 map to a) nor surjective
  (nothing maps to d).

80
Example()

• Which of the following sets P, Q, and R
  represents a partial, surjective, injective or
  bijective function from X1,2,3,4 to
  Ya,b,c,d? In each case explain your reasoning.
  
• Q lt1,cgt,lt2,agt,lt3,bgt,lt4,cgt,lt2,dgt
• Q is not a function from X to Y, because 2 maps
  to both a and d.

81
Example()

• Which of the following sets P, Q, and R
  represents a partial, surjective, injective or
  bijective function from X1,2,3,4 to
  Ya,b,c,d? In each case explain your reasoning.
  
• R lt1,cgt ,lt2,dgt,lt3,agt,lt4,bgt
• R is a total function from X to Y.
• Domain X, range Y. The function both injective
  and surjective.

82
Exercise

• Which of the following sets represents a
  functions from 1,2,3 to 4,5,6? If it is a
  functions, what type of function is it?.
• P lt1, 4gt ,lt2,5gt, lt3, 4gt
• Q lt1, 4gt ,lt2, 6gt
• R lt1, 4gt ,lt2, 1gt,lt3,3gt

83
Exercise

• Which of the following sets represents a
  functions from X 1,2,3,4 to Ya,b,c,d? If
  it is a functions, what type of function is it?
• P lt1,agt,lt2,agt,lt3,cgt,lt4,bgt
• Q lt1,cgt,lt2,agt,lt3,bgt,lt4,cgt,lt2,dgt
• R lt1,cgt ,lt2,dgt,lt3,agt,lt4,bgt
• R lt1,dgt ,lt2,dgt,lt4,agt
• R lt1,bgt ,lt2,bgt,lt3,bgt,lt4,bgt

84
Exercise-Solution 1

• Does P represents a functions from X 1,2,3,4
  to Ya,b,c,d? If it is a functions, what type
  of function is it?
• P lt1,agt,lt2,agt,lt3,cgt,lt4,bgt
• It is a function from X to Y.
• Domain X, range a,b,c. The function neither
  injective nor surjective.

85
Exercise-Solution 2

• Does Q represents a functions from X 1,2,3,4
  to Ya,b,c,d? If it is a functions, what type
  of function is it?
• Q lt1,cgt,lt2,agt,lt3,bgt,lt4,cgt,lt2,dgt
• It is not a function from X to Y.

86
Exercise-Solution 3

• Does R represents a functions from X 1,2,3,4
  to Ya,b,c,d? If it is a functions, what type
  of function is it?
• R lt1,cgt ,lt2,dgt,lt3,agt,lt4,bgt
• It is a function from X to Y.
• Domain X, range Y. The function both injective
  and surjective, therefore bijective.

87
Function on Sets

• Determine whether the set P represents a function
  from Xe,f,g,h to Ya,b,c,d.
• P lte,agt,ltf,agt,ltg,cgt,lth,bgt
• It is a function from X to Y.
• Domain X, range a,b,c.
• The function neither injective nor surjective.
  Why?

88
Function on Sets

• Determine whether the set Q represents a function
  from Xe,f,g,h to Ya,b,c,d.
• Qlte,cgt,ltf,agt,ltg,bgt,lth,cgt,ltf,dgt
• It is not a function from X to Y.
• Why?

89
Function on Sets

• Determine whether the set R represents a function
  from Xe,f,g,h to Ya,b,c,d.
• R lte,agt ,ltf,bgt,ltg,cgt,lth,dgt
• It is a function from X to Y.
• Domain X, range Y. The function both injective
  and surjective, therefore bijective.

90
Identifying Functions

• Which of the following are functions?
• Explain why they are or are not functions?
• parent(childPerson)-gtparentPerson
• mother(childPerson) -gt parentPerson
• oldestChild(parentPerson) -gt childPerson
• allChilderenOf(parentPerson) -gt ?(Person)

91
Identifying Functions()

• Let the sort Person represent the set of all
  people and the sort PPerson represent the power
  set of Person. Person lt PPerson
• Which of the following are total functions
• 1) Returns the parents of the argument
• op parentOf Person -gt Person
• 2) Returns the oldest child of the argument
• op oldestChildOf Person -gt Person
• 3) Returns all the children of the argument
• op allChildrenOf Person -gt PPerson

92
INJ Explained

• eq linv g f a a .
• ceq inj b0 b1 if g b0 g b1 .
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .
• Print out notes section slides 90-95, run slide
  show 90-95. Try to relate notes to slide show.

93
INJ Explained

• eq linv g f a a
• ceq inj b0 b1 if g b0 g b1
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .

Substitutions are f g b
b f g b
b
94
INJ Explained

• eq linv g f a a
• ceq inj b0 b1 if g b0 g b1
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .

giving ceq injNew f g b b if g f g b
g b
95
INJ Explained

• eq linv g f a a
• ceq inj b0 b1 if g b0 g b1
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .

Eval ceq injNew f g b b if g f g b
g b
evaluate
96
INJ Explained
Substitutions g b g b

• eq linv g f a a
• ceq inj b0 b1 if g b0 g b1
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .

giving ceq injNew f g b b if g f g b
g b
97
INJ Explained
Giving new equation eq linvNew
g f g b g b

• eq linv g f a a
• ceq inj b0 b1 if g b0 g b1
• open INJ .
• op b gt B .
• start f g b . right inverse
• apply .inj with b1 b at term
• apply red at term .

giving ceq injNew f g b b if g f g b
g b