1. Lecture 11 CS 1813 – Discrete Mathematics
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Lecture 11 CS 1813 – Discrete Mathematics
Algebra Every Which Way
Boolean Algebra
Predicate Algebra
Software Algebra
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

2. Algebraic Laws of Predicate Calculus
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Algebraic Laws of Predicate Calculus
x not free in q
(x. P(x))  (y. Q(y))
( )
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( )
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( )
( )
( )
y not in f(x)
(x. f(x)) = (y. f(y)) {R}
(x. f(x)) = (y. f(y)) {R}
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

3. Equational Reasoning with Predicates
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Equational Reasoning with Predicates
Theorem IX
( (x. P(x))  (y. Q(y)) ) = (x. y. P(x)  Q(y) )
Proof of Theorem IX
(x. P(x))  (x. Q(x))
= ((x. P(x)))  (x. Q(x)) {implication}
= (x. P(x))  (x. Q(x)) {3.6}
= x. ( (P(x))  (x. Q(x)) ) {3.10}
= x. ( (x. Q(x))  (P(x)) ) { comm}
= x. ( (y. Q(y))  (P(x)) ) {R}
= x. y. ( Q(y)  (P(x)) ) {3.10}
= x. y. ( (P(x))  Q(y) ) { comm}
= x. y. ( P(x)  Q(y) ) {implication}
qed
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

4. Equational Reasoning about Software
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Equational Reasoning about Software
Algebraic laws of sequence construction
(x: []) = [x] :[]
(xs  [ ] ) = (x. ys. xs = (x: ys) ) -- (:)
Informally
( x : [x1, x2, …] ) = [x, x1, x2, …] -- (: …)
Sequence length – an inductive definition
length(x: xs) = 1 + length xs (length).:
length[ ] = (length).[]
Theorem (len0). xs::[a]. length xs  0
We are going to accept this theorem without proof
Think of it as axiomatic
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

5. Zero Length Means Empty
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
Zero Length Means Empty
1/2/2019
length :: [a] -> Int
length(x: xs) = 1 + length xs (length).:
length[ ] = (length).[]
Theorem (len 0  [ ]). length xs = 0  xs = [ ]
Proof, part 1 (): xs = [ ]  length xs = 0
length xs
= length [ ] assumption: xs = [ ]
= (length).[]
Proof, part 2 (): xs  [ ]  length xs  0 (contrapositive)
= length(x: ys) (:) - (x. ys. xs = (x: ys))
= 1 + length ys (length).:
 (len0)
= nd grade arith
 nd grade arith
Corollary (:len). length xs  0  xs  [ ]  x. ys. (xs = (x: ys))  ((length xs) = ((length ys) + 1))
We’ll use this theorem
many, many times
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

6. Reasoning about Inductive Equations
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Reasoning about Inductive Equations
sum :: Num a => [a] -> a
Two equations any good sum function should satisfy
sum(x: xs) = x + sum xs (sum).:
sum[ ] = (sum).[]
Pleasant surprise
These two equations are enough to completely define sum
Just use equational reasoning
sum[7, 19, -12] = sum(7: [19, -12]) (:)
= 7 + sum([19, -12]) (sum).:
= 7 + sum(19: [-12]) (:)
= 7 + (19 + sum([-12])) (sum).:
= 7 + (19 + sum(-12: [ ])) (:)
= 7 + (19 +((-12) + sum[ ])) (sum).:
= 7 + (19 + ((-12) + 0)) (sum).[]
= nd-grade arithmetic
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

7. But What About sum[x1, x2, … xn]?
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
But What About sum[x1, x2, … xn]?
Theorem proved
sum[7, 19, -12] = 14
Theorem that could have been proved with essentially the same argument
Universe of discourse for addends: a is a type of class Num
x1, x2, x3 :: a . sum[x1, x2, x3] = x1 + (x2 + (x3 + 0))
Theorem that should be proved
nN Note: N = {0, 1, 2 …}
xk :: a . sum[x1, x2, … xn] = x1+ (x2 + (x3 + … (xn-1 + (xn + 0)) …))
For any fixed value of n, reason as before
Proof will have 2n+1 steps
But, we want to prove it for all possible values of n, not just those for which we have time to do a proofs with 2n+1 steps
Conclusion
Relying on a new proof for each n is not good enough
Must find a better way
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

8. The Principle of Mathematical Induction
Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
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)
THE SINGLE MOST IMPORTANT CONCEPT YOU WILL LEARN THIS YEAR, NEXT YEAR, OR THE YEAR AFTER THAT
principle of
induction
This is a subtle idea. Only P(0) is proven directly.
The rest gives a template for constructing a proof of P(1) or P(1000)
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

9. Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
sum works!
P(0) n.P(n)P(n+1)
{Ind}
n. P(n)
Induction
Principle of Induction
Prove: P(0)
Prove: P(n)  P(n+1), n arbitrary
Conclude: nN. P(n)
Define P(n) to be the following proposition
sum[x1, x2, … xn] = x1+ (x2 + (x3 + … (xn-1 + (xn + 0)) …))
What about P(0)?
P(0) says this: sum[ ] = 0
True because that is what the equation (sum).0 says
Now prove the implication P(n)  P(n+1)
The implication is always True when P(n) is False
So, we only need to prove that P(n+1) is True when P(n) is True
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

10. Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
Coup de Grâce
P(n +1) is the proposition sum[x1, x2, … xn+1] = x1+ (x2 + (x3 + … (xn + (xn+1 + 0)) …))
sum[x1, x2, … xn+1]
= sum(x1 : [x2, … xn+1]) (: …)
= x1 + (sum[x2, … xn+1]) (sum).:
= x1 + (x2 + (x3 + … (xn + (xn+1 + 0)) …)) P(n)
That is sum[x1, x2, … xn+1] = x1 + (x2 + (x3 + … (xn + (xn+1 + 0)) …))
But, that is exactly what P(n+1) says
This proves the implication P(n)  P(n+1)
Conclude nN. P(n) — by the Principle of Induction
That is: nN. sum[x1, x2, … xn] = x1+ (x2 + (x3 + … (xn-1 + (xn + 0)) …))
qed
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

11. Lecture 12 - CS 1813 Discrete Math, University of Oklahoma
1/2/2019
End of Lecture
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page