CSCE 235
Homework
Assignment 1 – Solution
Assigned: January 12, 2009
Due: 12:30 p.m. January 21,
2009
(Homework 5 minutes late will not be
accepted)
Logic Basics
1. (5 points) Consider the following propositions:
, 
, 
, 
, 
, 
, 
, 
,
(a) Which
proposition is the converse of 
? (Identify all equivalences)
,
(b) Which
proposition is the contrapositive of
? (Identify all equivalences)
,
2. (5 points) Suppose that 
is known to be
false. Give the truth values for
(a) 
(b) 
(c) 
If is known to be false,
that means when p is true, q is false, as that is the only
condition that could cause to be false. Thus, we can conclude that when p is
true, q is false. So:
(a) is the AND of true and false. Thus is false.
(b) is the OR of true and false. Thus is true.
(c) is 
. Now, since p
is true, is true.
3. (5 points) Construct the truth table for
(a) 
(b) 
(c) 
(d) 
(e) 
(a)
|
p |
q |
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0 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
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0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
|
1 |
0 |
1 |
0 |
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
|
Step |
1 |
4 |
2 |
1 |
3 |
1 |
|
(b)
|
p |
q |
r |
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0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
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0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
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0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
|
1 |
0 |
0 |
1 |
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
|
Step |
1 |
2 |
3 |
2 |
1 |
||
(c)
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p |
q |
r |
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0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
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0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
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0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
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0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
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1 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
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1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
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Step |
1 |
2 |
1 |
3 |
1 |
2 |
||
(d)
|
p |
q |
r |
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0 |
0 |
0 |
1 |
0 |
0 |
0 |
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0 |
0 |
1 |
1 |
0 |
1 |
1 |
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0 |
1 |
0 |
0 |
1 |
1 |
0 |
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0 |
1 |
1 |
0 |
1 |
1 |
1 |
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1 |
0 |
0 |
0 |
1 |
1 |
0 |
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1 |
0 |
1 |
0 |
1 |
1 |
1 |
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1 |
1 |
0 |
0 |
1 |
1 |
0 |
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1 |
1 |
1 |
0 |
1 |
1 |
1 |
|
Step |
2 |
1 |
3 |
1 |
||
(e)
|
p |
q |
r |
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(( |
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0 |
0 |
0 |
0 |
0 |
1 |
0 |
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0 |
0 |
1 |
0 |
0 |
1 |
1 |
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0 |
1 |
0 |
1 |
1 |
0 |
0 |
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0 |
1 |
1 |
0 |
1 |
1 |
1 |
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1 |
0 |
0 |
1 |
1 |
0 |
0 |
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1 |
0 |
1 |
0 |
1 |
1 |
1 |
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1 |
1 |
0 |
1 |
1 |
0 |
0 |
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1 |
1 |
1 |
0 |
1 |
1 |
1 |
|
Step |
3 |
1 |
2 |
1 |
||
)
Logic Equivalences and Implications
4. (5 points) Verify the following logical implications using truth tables.
(a) 
, addition
(b) 
, disjunctive syllogism
(a) To
prove , the new must show that is a tautology.
|
p |
p |
|
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0 |
0 |
1 |
0 |
|
1 |
1 |
1 |
1 |
|
Step |
0 |
2 |
1 |
Since the final column (Step 2) has all truth values true,
we show that is a tautology. Thus, we have proven .
(b) To prove , the new must show that 
is a tautology.
|
p |
q |
|
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0 |
0 |
0 |
0 |
1 |
1 |
0 |
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0 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
0 |
1 |
1 |
|
Step |
1 |
2 |
1 |
3 |
1 |
|
Since the final column (Step 3) has all truth values true,
we show that is a tautology. Thus, we have proven .
5. (10 points) Prove or disprove the following. (Hint: only one line of the truth table is needed to show that a proposition is not a tautology.)
(a) 
(b) 
(c) 
(d) 
(e) 
(a)
|
p |
q |
|
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1 |
0 |
1 |
0 |
0 |
We show in the above that when p is true and q is
false, 
is false. Therefore, is false.
(b)
|
p |
q |
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1 |
0 |
1 |
1 |
0 |
0 |
We show in the above that when p is true and q is
false, 
is false. Therefore, is false.
(c)
|
p |
q |
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0 |
0 |
0 |
1 |
0 |
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0 |
1 |
0 |
1 |
1 |
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1 |
0 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
1 |
|
Step |
1 |
2 |
1 |
|
We show in the above that 
is a tautology. Therefore, is true.
(d) 
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p |
q |
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0 |
0 |
0 |
0 |
1 |
0 |
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0 |
1 |
0 |
0 |
1 |
0 |
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1 |
0 |
1 |
0 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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Step |
2 |
1 |
3 |
1 |
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We show in the above that 
is a tautology. Therefore, is true. This is actually known as the Absorption Law.
(e)
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p |
q |
r |
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0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
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0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
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0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
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0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
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1 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
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1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
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1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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Step |
1 |
2 |
1 |
2 |
1 |
||
We show in the above that 
is a tautology. Therefore, is true.
This is actually known as the Exportation Law.
6. (5
points) The
“exclusive or” connective 
is defined by the
truth table:
|
p |
q |
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0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
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1 |
1 |
0 |
(a) Show
that 
.
(b) Show that 
.
(a) To show that:
|
p |
q |
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0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
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0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
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1 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
|
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
|
Step |
1 |
4 |
1 |
3 |
2 |
1 |
|
We show in the above that 
is a tautology. Therefore, is true.
(b) Show that .
|
p |
q |
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0 |
0 |
0 |
1 |
0 |
1 |
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0 |
1 |
1 |
1 |
1 |
0 |
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1 |
0 |
1 |
1 |
1 |
0 |
|
1 |
1 |
0 |
1 |
0 |
1 |
|
Step |
1 |
3 |
2 |
1 |
|
We show in the above that 
is a tautology. Therefore, is true.
7. (10 points) Every compound
proposition can be written using only the connectives 
and 
. This fact follows
from the equivalences 
, 
, and 
. Find propositions
logically equivalent to the following using only the connectives and .
(a) 
(b) 
(c) 
(d) 
(Hint: see problem
6(b).)
(a) 
Definition
of Equivalence
Implications
Double
Negation
De Morgan
(b) 
Implications
Double
Negation
De Morgan
(c) 
Implications
Double
Negation
De Morgan
(d) (Hint: see problem
6(b).)
We see that . So
Equivalence
De
Morgan
Implications
8. (10 points) The Sheffer Stroke is a connective 
defined by the truth table:
|
p |
q |
|
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0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
This connective is interesting because all compound propositions can be written using only this connective. You may prove logically or use a truth table (or a combination of both) for the following.
(a) Show
that 
.
(b) Show that 
.
(c) Find
a proposition equivalent to using only the Sheffer Stroke.
(a) Show
that . First we show using
a truth table
|
p |
|
|
|
|
0 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
|
Step |
1 |
2 |
1 |
Since 
has
all truth values true, it is a tautology.
Therefore, is true.
Now, we want to show it using logic. First, let us look a the
Sheffer Stroke’s truth table. For each line that results
in a true value (i.e., 1), we can write a proposition. So we have
.
Each conjunction is for a line of the truth table that
results in a true value. Now, we can
simplify the above proposition.
Distributive
Negation
Identity
Distributive
Negation
Identity
So, we have shown that 
; and thus we show that 
. Now, for 
, we have 
. Therefore, we show
that is true.
(b) Show
that . First we show using a truth table.
|
p |
q |
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0 |
0 |
0 |
1 |
1 |
0 |
1 |
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0 |
1 |
1 |
1 |
1 |
1 |
0 |
|
1 |
0 |
1 |
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
0 |
1 |
0 |
|
Step |
1 |
3 |
1 |
2 |
1 |
|
We show in the above that 
is a tautology. Therefore, is true.
Now, we want to show it using logic. We have previously shown that
. So, we have
Using the result from (a).
De
Morgan
Therefore, we show that is true.
(c)
Find a proposition
equivalent to using only the Sheffer Stroke. We
have:
Implication
Previously, in (b), we have 
. Thus, 
. Or, to simplify it
further, since, in (a) and 
, we have 
. Thus, can be expressed using
on the Sheffer Stroke: 
.
Formal Proofs
9. (15 points) Complete the following formal proofs by supplying explanations for each step.
(a) If 
, 
and 
, then 
.
Proof Explanations
1. Given, or Hypothesis
2. Given,
or Hypothesis
3. Given,
or Hypothesis
4. 
2;
implication
5. 
4;
commutative
6. 
1,
5; modus tollens
7. 3,
6; disjunctive syllogism
(b) If 
, 
, 
, then 
.
Proof Explanations
1. Hypothesis
2. Hypothesis
3. Hypothesis
4. 
2,3; Hypothetical Syllogism
5. 
1; Simplification
6. 
4; Simplification
7. 
4; Simplification
8. 
7; Double Negation
9. 
4,
8; Modus Tollens
10. 9,
6; Conjunction
(c) If 
, 
and 
, then 
.
Proof Explanations
1. Given,
or Hypothesis
2. Given, or Hypothesis
3. Given,
or Hypothesis
4. 
3;
Addition
5. 
4;
Implication
6. 1,
5; Modus Tollens
7. 6;
Double Negation
8. 
2,
7; disjunctive syllogism
10. (20 points) Complete the following proof by contradiction by supplying explanations for each step.
(a) If , , , then .
Proof Explanations
1. Given,
or Hypothesis
2. Given,
or Hypothesis
3. Given,
or Hypothesis
4. 
Negation
of conclusion
5. 
4;
De Morgan
6. 
5;
Commutative
7. 
6;
Implication
8. 
7,
2; transitivity of hypothetical syllogism
9. 
8,
3; transitivity of hypothetical syllogism
10. 
1;
Simplification
11. 1;
Simplification
12. 11;
Simplification
13. 11;
Simplification
14. 12,
9; Modus Ponens
15. 
14,
3; Conjunction
16. contradiction 15; Negation
(b) If
, 
, , and 
, then .
Proof Explanations
1. Hypothesis
2. Hypothesis
3. Hypothesis
4. Hypothesis
5. 
Negation
of conclusion
6. 
5,
6; Conjunction
7. 
6; Double Negation
8. 
7; De Morgan’s
9. 
8; Contrapositive
10. 2,
9; Modus Tollens
14. 
10,
3; Conjunction
15. 
1,
14; Conjunction
16. contradiction 15; Negation
•
Based on Ross and Wright (1988).