1. Quizzes CS 1813 – Discrete Mathematics
Lecture 7 - CS 1813 Discrete Math, University of Oklahoma
12/2/2018
Quizzes CS 1813 – Discrete Mathematics
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

2. CS 1813 Discrete Mathematics, Univ Oklahoma
But First … a quiz
Prove the following sequent
a  (b) |– (a  b)
Proof
{I} triggers discharge
a  (b False)
{EL} a
a  b
a  (b False)
{E} b
{ ER}
b False
{E}
False
{I}
(a  b)  False
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

3. CS 1813 Discrete Mathematics, Univ Oklahoma
But First … a quiz
Prove the following sequent
|– (a  b)  (b  c)
Proof
{I} triggers discharge
a  b
{ ER} b
{IL}
b  c
{I}
a  b  b  c
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

4. CS 1813 Discrete Mathematics, Univ Oklahoma
But First … a quiz
Prove the following Boolean equation
(a  b) = a  (b)
Proof
(a  b)
= ((a)  b) {implication}
= ((a))  (b) {DeMorgan }
= a  (b) {double neg}
QED
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page

5. CS 1813 Discrete Mathematics, Univ Oklahoma
Quiz
map f [ ] = [ ] (map).[]
map f (x: xs) = (f x): (map f xs) (map).:
Prove
k  {1, 2, …, (length xs)}. (map f xs)k = f(xsk)
ysk denotes the kth element of ys
for any sequence ys = [y1, y2, … yn], [y1, y2, … yn]k = yk
Assume that xs has a finite number of elements
Proof
k  {1, 2, …, 0}. (map f xs)k = xsk (Isaac’s rule, since {1, …, 0} = )
k  {1, 2, …, (length(x: xs))}. (map f (x: xs))k = f((x: xs)k) = k  {1, 2, …, 1+(lengthxs)}. ((f x): (map f xs))k = f((x: xs)k)
True for k=1
((f x): (map f (x: xs)))1 = (f x) = f((x: xs)1)
True for 2  k  1+(length xs)
induction hypothesis, since (y: ys)k = ysk-1 when k  2
CS Discrete Mathematics, Univ Oklahoma
Copyright © 2000 by Rex Page