CS 1813 – Discrete Mathematics
Review of
Predicate Calculus
Set Theory
Inductive Definition
Proof by Mathematical Induction
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
1
What is a Predicate?
Predicate
 Parameterized collection of propositions
 P(x)
 Typically a different proposition for each x
 Universe of discourse
– Values that x may take
Universe of discourse
 Must be specified
 Otherwise, all bets off — muchas contradicciónes
 Non-empty
 Empty universe calls for special handling
 Default assumption: non-empty universe
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
2
 — the Universal Quantifier, Forall
 x.P(x)
 This formula is a WFF of predicate calculus whenever P(x) is a
WFF of predicate calculus
 True if the proposition P(x) is True for all x’s in the universe of
discourse
 False if there is some x in the universe of discourse for which
P(x) is False
 Equivalent to forming the Logical And of all P(x)’s
 Example – S predicate about sum
 S(n)  sum[x1, x2, …, xn] = x1 + x2 + … + xn
 n.S(n)
 Universe of discourse: natural numbers N = {0, 1, 2, … }
 n.S(n) means S(0)  S(1)  S(2)  …
 So, “” provides a way to write formulas that would contain
an infinite number of symbols if written in propositional
calculus notation (but infinitely long formulas aren’t WFFs)
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
3
 —the Existential Quantifier, There Exists
 x.P(x)
 This formula is a WFF of predicate calculus whenever P(x) is a
WFF of predicate calculus
 True if there is at least one x in the universe of discourse for
which the proposition P(x) is True
 False if x. P(x) is True
 Equivalent to forming the Logical Or of all P(x)’s
 Example – E predicate about maximum
 E(n, k)  maximum[s1, s2, …, sn] = sk
 k.E(23, k)
 Universe of discourse: U = {1, 2, …, 23}
 k.E(23, k) means E(23,1)  E(23,2)  …  E(23,23)
 Note: When U is finite, quantifiers not required
– Clumsy to write big formulas without quantifiers, though
– Without quantifiers, reasoning can be more complex, too
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
4
Free Variables and Bound Variables
Variables in WFFs of predicate calculus
 Denoted by lower-case letters
 Examples of predicate calculus WFFs with variables
 F(p, q)  G(q, r)
 (x.F(x))  (G(y)  H(y))
 (x.F(x, y)  G(y))  (H(z)  K(x))
variables: p, q, r
variables: x, y
variables: x, y, z
Free variables and bound variables
 Let e stand for a WFF of predicate calculus
 Bound variable
 x. e
 x. e
x is bound in the formula x. e
x is bound in the formula x. e
 Free variables are variables that are not bound
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
5
Arbitrary Variables
 A variable is arbitrary in a proof if it does not occur
free in any undischarged assumption of that proof
 Examples
discharged
x. F(x)
{E}
F(x)
x arbitrary? Yes
G(x, y)
{I}
y. G(x, y)
x arbitrary? No
P(x)  Q
{EL}
P(x)
x arbitrary? Yes
{I}
P(x)  Q  P(x)
{I}
x. P(x)  Q  P(x)
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
6
Inference Rules of Predicate Calculus
Renaming Variables
F(x) {x, y arbitrary, y not in F(x)}
{R}
F(y)
x. F(x)
{y not in F(x)}
{R}
y. F(y)
x. F(x)
{y not in F(x)}
{R}
y. F(y)
Introducing/Eliminating Quantifiers
F(x) {x arbitrary}
x. F(x) {universe is not empty}
{I}
{E}
F(x)
x. F(x)
F(x)
{I}
x. F(x)
x. F(x) F(x) |– A {x not free in A}
{E}
A
...plus the inference rules
{E} rule triggers discharge
of propositional calculus
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
7
Existential Elimination
something like {E}
Theorem 31
x. P(x), x. P(x)Q(x) |– x. Q(x)
proof
x. P(x)Q(x)
{E}
P(x) P(x)Q(x)
P plays role of F
{E}
in {E} rule
Q(x)
{I}
x. P(x)
x. Q(x)
{E}
x. Q(x) {x not free in x.Q(x)}
x. F(x) F(x) |– A {x not free in A}
{E}
A
xU. Q(x) plays
role of A in {E} rule
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
8
Existential Incrimination
it’s not legal, really
Theorem 9 from Planet X: x. P(x), x. P(x)Q(x) QED
|– x. Q(x)
Proof
Why not?
 Suppose x U, where U is the domain of discourse for x
 Is P(x)Q(x) True?
 Maybe so, maybe not
 “x. P(x)Q(x)” means P(a)Q(a) is True for at least one value a U
 There may be many values b  U where P(b)Q(b) is False
 Counter-example that disproves the theorem
 Universe of discourse = {0, 1}
 P(0) = True, P(1) = True
Note: (x. P(x)) = True
 Q(0) = True, Q(1) = False
Note: (x. P(x)Q(x)) = True
 But, (x. (P(x)Q(x))) is also True
 Because P(1) Q(1) is False
 (x. Q(x)) = False, because Q(1) = False
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
9
Algebraic Laws of Predicate Calculus
(x. P(x))  (y. Q(y))
(
(
(
(
(
(
)
)
)
)
)(
) (
)
)
(
) (
(x. f(x)) = (y. f(y))
(x. f(x)) = (y. f(y))
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
)
{R}
{R}
10
Equational Reasoning with Predicates
Theorem
( (x. P(x))  (x. Q(x)) ) = (x. y. P(x)  Q(y) )
 Proof of Theorem
(x. P(x))  (x. Q(x))
= ((x. P(x)))  (x. Q(x))
= (x. P(x))  (x. Q(x))
= x. ( (P(x))  (x. Q(x)) )
= x. ( (x. Q(x))  (P(x)) )
= x. ( (y. Q(y))  (P(x)) )
= x. y. ( Q(y)  (P(x)) )
= x. y. ( (P(x))  Q(y) )
= x. y. ( P(x)  Q(y) )
{implication}
{3.6}
{3.10}
{ comm}
{R}
{3.10}
{ comm}
{implication}
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
qed
11
Equational Reasoning about Software
 Algebraic law of sequence construction
x : [x1, x2, …, xn] = [x, x1, x2, …, xn]
-- (:)
(++) :: [a] -> [a] -> [a]
[ ] ++ ys = ys
(x : xs) ++ ys = x : (xs ++ ys)
-- (++).[]
-- (++).:
 Algebraic laws of concatenation
 An equational argument
[x] ++ xs
= (x : [ ]) ++ xs
= x : ([ ] ++ xs)
= x : xs
 What did this prove?
Theorem (++1)
[x] ++ xs = x : xs
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
(:)
(++).:
(++).[]
()
12
More Software Equations
 Algebraic laws of foldr
foldr (the big picture)
foldr () z [x1, x2, …, xn] =
x1  (x2  … (xn-1  (xn  z)) … )
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr () z [ ] = z
foldr () z (x : xs) = x  (foldr () z xs)
 The big or
-- (foldr).[]
-- (foldr).:
(\/) :: Bool -> Bool -> Bool -- “little or” – satisfies Boolean laws for 
or :: [Bool] -> Bool
-- “big or”
or = foldr (\/) False
-- (or)
 Theorem (or null1) — or null/first element
or ([True] ++ xs) = True
or ([True] ++ xs)
= or (True : xs)
= foldr (\/) False (True : xs)
= True \/ (foldr (\/) False xs)
= True
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
(++1)
(or)
(foldr).:
( null)
13
qed
The Principle of
Mathematical Induction
Universe of discourse: N = {0, 1, 2, …}
Predicate P
 P(n) is a proposition whenever nN
 Want to prove: nN. P(n)
 That is, to prove that the proposition P(n) is True for
all natural numbers n
Principle of Induction
 Prove: P(0)
 Prove: nN. (P(n)  P(n+1))
 Conclude: nN. P(n)
principle
of
induction
A way to prove P(n) for all values of n,
without proving any P(n) directly except P(0)
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
14
Theorem — or[x, y, …,True, …] = True
 Theorem (orT)
nN. length ys = n  or (ys ++ ([True] ++ xs)) = True
 Proof
 P(n)  length ys = n  or (ys ++ ([True] ++ xs)) = True
 Base case: P(0)  length ys = 0  or(ys ++ ([True] ++ xs)) = True
length ys = 0
 ys = [ ]
zero len theorem
 or(ys ++ ([True] ++ xs)) = or([ ] ++ ([True] ++ xs)) substitution
= or ([True] ++ xs)
++.[]
= True
or1T
 Inductive case: P(n)  P(n+1)
… next slide
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
15
Theorem — or[x, y, …,True, …] = True
Inductive Case
 Predicate to prove for inductive case
 P(n+1)  length ys = n+1  or (ys ++ ([True] ++ xs)) = True
length ys = n+1
 ys  [ ]
zero len theorem
 ys = y: zs  length zs = n
:len corollary
 or(ys ++ ([True] ++ xs)) = or((y:zs) ++ ([True] ++ xs))
subst
 length zs = n
= or(y:(zs ++ ([True] ++ xs)))  length zs = n
= y \/ (or(zs ++ ([True] ++ xs))))  length zs = n
= (y \/ True)  length zs = n
induction hypothesis
 (y \/ True)
= True
 Conclude: nN. P(n)
(++).:
(foldr).:
P(n)
(conj imp)
 null
principle of induction
CS 1813 Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page
qed
16
Strong Induction
n.(mn.P(m))P(n)
Basic Principle of Induction
{StrInd}
 Prove P(0) is true
n.P(n)
 Prove P(n+1) is true, nN arbitrary
Strong Induction
 Proof of P(n+1) may assume P(n) is true
 Conclude nN. P(n)
 What if proof of P(n+1) does not make use of P(n)?
 Conclusion nN. P(n) is still valid
 This would be a direct proof using the {I} inference rule
 The principle of induction is a lever that makes the proof easier
 Principle of Strong Induction
 Prove P(n) for arbitrary n  N
 Twist: Can assume kDn.P(k), where Dn= {kN | k  n}
 Conclude nN. P(n)
Pretty good deal, eh?
 Getting Started: D0 = { }
 So, must prove P(0) from scratch (as with old principle of induction)
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Copyright © 2000 by Rex Page
17
msort Termination Proof — inductive case
msort (x1: x2: xs) = merge (msort ys) (msort zs)
where (ys, zs) = deal(x1: x2: xs)
(msort).:
 P(n)  (length xs) = n  (msort xs) terminates  length(msort xs) = n
 Proof of P(n) for arbitrary n  {2, 3, 4, …}
merge terminates (mT)
msort xs
= msort(x1: x2: ws)
(pattern matching) (length xs)  2
= merge (msort ys) (msort zs) (no operation)
(msort).:
deal terminates (dS)
where (ys, zs) = deal(x1: x2: ws)
(msort ys) terminates – Why? Ind Hyp implies
P(length ys) is True
strong induction
(msort zs) terminates (similar argument)
Each step (matching, deal, msort twice, merge) takes a finite amount of time,
so (msort xs) terminates
 length xs = length ys + length zs
(dS)
= length(msort ys) + length(msort zs)
P(length ys), P(length zs)
= length(merge (msort ys) (msort zs))
(mT)
= length(msort xs)
(msort).:: – as in above proof
length ys  length xs (dR)
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Copyright © 2000 by Rex Page
18
qed (strong induction)
Haskell Type Specifications
 x, y, z :: Integer
 xs, ys :: [Integer]
 or :: [Bool] -> Bool
 (++) :: [e] -> [e] -> [e]
 sum :: Num n => [n] -> n
-- x, y, and z have type Integer
-- sequences with Integer elements
-- function with one argument
argument is sequence with Bool elems
delivers value of type Bool
-- generic function with two arguments
args are sequences with elems of same type
type is not constrained (can be any type)
delivers sequence with elements of
same type as those in arguments
-- generic function with one argument
argument is a sequence with elems of type n
n must a type of class Num
Num is a set of types with +, , … operations
 powerSet :: (Eq e, Show e) => Set e -> Set(Set e)
-- generic function with one argument
argument is a set with elements of type e
delivers set with elements of type (Set e)
type e must be both class Eq and class Show
Class Eq has == operator, Show displayable
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Copyright © 2000 by Rex Page
19
Inductive Definitions
aka Recursive Equations
sum :: Num n => [n] -> n
sum(x: xs) = x + sum xs
sum[ ] = 0
Theorem: sum = foldr (+) 0
length :: [a] -> Int
length(x: xs) = 1 + length xs
length[ ] = 0
Theorem: length = foldr oneMore 0
where oneMore x n = 1 + n
(sum).:
(sum).[]
(length).:
(length).[]
(++) :: [a] -> [a] -> [a]
(x: xs) ++ ys = x: (xs ++ ys)
[ ] ++ ys = ys
Theorem: xs ++ ys = foldr (:) ys xs
(++).:
(++).[]
concat :: [[a]] -> [a]
concat(xs: xss) = xs ++ concat xss
concat[ ] = [ ]
Theorem: concat = foldr (++) [ ]
(concat).:
(concat).[]
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Copyright © 2000 by Rex Page
20
Patterns of Computation
Pattern: foldr () z [x1, x2, …, xn-1, xn ] = x1  ( x2  … (xn-1  ( xn  z)) …
)
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr () z (x: xs) = x  foldr () z xs
(foldr).:
foldr () z [ ] = z
(foldr).[]
Pattern: map f [x1, x2, … xn] = [f x1, f x2, … f xn]
map :: (a -> b) -> [a] -> [b]
map f (x : xs) = (f x) : map f xs
map f [ ] = [ ]
(map).:
(map).[]
Pattern: zipWith b [x1, x2, … xn] [y1, y2, … yn] =
[b x1 y1, b x2 y2, … b xn yn]
Note: extra elements in either sequence are dropped
zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith b (x:xs) (y:ys) = (b x y): (zipWith b xs ys) (zipW).:
zipWith b [ ] ys = [ ]
(zipW).[]-1
zipWith b xs [ ] = [ ]
(zipW).[]-2
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Copyright © 2000 by Rex Page
21
 {2, 3, 5, 7, 11}
 2  {2, 3, 5, 7, 11}
Sets
— explicit enumeration
— stylized epsilon means “element of”
 ={}
 {x | p x}

— stylized Greek letter phi denotes empty set
— set comprehension
 Denotes set with elements x, where (p x) is True
{f x | p x}
— set comprehension
 Denotes set with elements of form (f x), where (p x) is True
A  B  x. (x  A  x  B)
— subset
A = B  (A  B)  (B  A)
— set equality
A  B = {x | x  A  x  B}
— union
S = {x | AS. x  A}
— big union
A  B = {x | x  A  x  B}
— intersection
S = {x | AS. x  A}
— big intersection
A – B = {x | x  A  x  B}
— set difference







 A’ = U – A
 P(A) = {S | S  A}
 A  B = {(a, b) | a  A  b  B}
— complement (U = universe)
— power set
— Cartesian product
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22
End of Lecture
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Copyright © 2000 by Rex Page
23