RAYMOND SMULLYANS LOGICAL PUZZLES
MATH 320, SPRING 2014
On an island live two kinds of inhabitants, knights who always tell the truth and, and their opposites, knaves, who
always lie. You encounter two people A and B . What are A and B if A says B is a kn
Homework 2

(Your name)
Math. 320, Spring 2014
IMPORTANT
SHOW ALL YOUR WORK
:
Please do all your work in space provided.
If needed, you can use backspaces. No additional sheets of paper will be accepted.
Check that your homework has a total of 5 pagesth
Homework 1
Math. 320, Spring 2014
SHOW ALL YOUR WORK
IMPORTANT: Please do all your work in space provided.
If needed, you can use backspaces. No additional sheets of paper will be accepted.
Check that your homework has a total of 6 pagesthere are no blan
Math. 320, Spring 2014
Examples of proofs by induction
Example 1
(Bernoulli's inequality).
! 1
For all x
and n
!1+
(1 + x)
n
Proof by induction.
PN
,
(1)
nx:
For n = 1, we have
(1 + x)1 = 1 + x ! 1 + 1 x = 1 + x:
Suppose (1) is true for n = k. We have
(1
Math. 320, Spring 2014
Each inhabitant of a remote village always tells the truth or always lies. A villager will only give a
Yes or a No response to a question a tourist asks. Suppose you are a tourist visiting this area and come
to a fork in the road. O
Calculus and Logic
Math. 320 Spring 2014
Example 1: Use quantiers to express the denition of the limit of a realvalued function f (x) at a point
a, with D being a domain of f .
Solution: Recall the denition of the statement
lim f (x) = L
xa
is For every
Math. 320, Spring 2014
A few comments and examples on proofs
Direct proofs
A direct proof of the statement P Q is based on two rules of inference:
[p (p q )] q
[(p q ) (q r)] (p r)
(Modus ponens)
(Hypothetical syllogism)
It may be fairly easy to prove the
Math. 320, Spring 2014
The proofs presented below, in part, are based on the material from Discrete Mathematics, by
R. C. Penner, World Scientic, 1999, and Discrete Structures, Logic, and Computability, by
J. L. Hein, Jones and Bartlett Publishers, 2002
D
Examples of Disjunctive Normal Forms (DNF)
Math. 320 Spring 2014
p p
1
0
=
0
p p, i.e., p is already in its DNF.
1
p q pq
11
1
10
1
01
1
00
0
=
pq
(p q ) (p q ) (p q ) p (q q ) (p q )
p (p q ) p q .
In other words, p q is already in its DNF.
p q pq
11
1