Notes on Assignment 18: p243, 1, 2acdfgij, 3, 4, [5e-k was moved into Homework 14]
Notes on problem 1: You have used truth tables to show whether two formulas are
equivalent or not. This question was different: youre asked whether the truth of one
formula
Quiz 3 Review.doc
Ling 409, December 7, 2005
Review for Quiz 3: Algebra, Automata, and Grammars
Quiz 3 is a closed book quiz except for one handout with axioms for algebras, identical to
a one-page handout I gave you during the algebra unit.
Algebra will
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Homework 18 p.243-4: 1,2acdfgij;3;4;5e-k
(5e) No ducks are amphibious
x(D(x) ~A(x)
or ~x(D(x) & A(x)
(5f) Every cloud has a silver lining
x(C(x) y(L(y) & H(x,y)
L(y) = y is a silver lining
H(x,y) = x has y
(5g) Only Rosicrucians experience complete happ
Answer to a Mathematical Induction Problem
PtMW, Chapter 8, Exercise 4, p.233
4. Prove by induction that the power set of a set with n members has 2n members, for any
finite positive n. You do not have to give the proof in formal steps that look like a lo
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Hmwk 11: pp.139-134: 9a-d; 10 (choose any 3); 11; 13ai; 13bi
(9b)
1. p
2. r
3. (p & r)
4. (p & r) q
5. q
conj.1,2
3,4: Modus Ponens
Notes: Modus Ponens says that given PQ and P, then Q.
Since p, r, then (p & r).
Therefore, by Modus Ponens, q.
(9c)
1. p
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pp.129-134, 1-3, 4a-c, 5a-c, 6a-d, 7, 8a-c
Wednesday, October 03, 2001
Many of the answers are in the back of the book. Here I provide working for (most of)
them, plus comments on how to approach them.
(3a) (p&q) & s)
= (T&T)&F
=T&F
=F
(3c) ps
= TF
=F
(
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Homework 5: pp.51-3, #4,5
Answers
(4) Faulty reasoning:
A relation R is reflexive if for all xdom(R), <x,x>R.
The reasoning does not allow for the case that there is some xdom(R) s.t. for all
yrange(R), <x,y>R (i.e. x bears no relation in R).
For exampl
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Homework 4: pp.51-2, 1,2,3
Questions
Q1: Whats the difference between antisymmetric and asymmetric?
A1:
R is asymmetric if for all <x,y> R, <y,x> R.
R is antisymmetric if for all <x,y> R, <y,x> R unless x=y.
e.g. cfw_<1,1>, <1,2>, <2,3> is not asymmetri
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Homework 3
pp.36-7: 1,2,3a
(1)
(a)
(i) cfw_<b,2>, <b,3>, <c,2>, <c,3>
(ii) cfw_<2,b>, <2,c>, <3,b>, <3,c>
(iii) cfw_<b,b>, <b,c>, <c,b>, <c,c>
(iv) cfw_<b,2>, <b,3>, <c,2>, <c,3>, <2,2>, <2,3>,<3,2>,<3,3>
(v) , since A B = .
(vi) Same as A B
(b)
(i) Tru
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Homework 2: Supplemental Answers
Paul de Lacy
Monday, September 10, 2001
(9a) I have expanded this each detail at a time for the sake of clarity.
There are several ways to prove this problem; this is only one.
(A C) ( BC) (A B)
(A C) ( BC) ( A B) = (A B
Homework 1 Supplementary Answers and Notes
Paul de Lacy
Friday, September 07, 2001
Questions Asked in Class and in the Homeworks
Q: Is a cfw_a?
A: No. cfw_a cfw_a.
The confusion here is with the notion member. Informally, x is a member of a
set if it appe