JDEP183H
Handout
25b: Some Exercises in Propositional
Logic -- Solution
October 14, 2006
Logic Basics
1. Consider the following propositions:
, 
, 
, 
, 
, 
, 
, 
(a) Which
proposition is the converse of ?
(b) Which
proposition is the contrapositive of
?
(c) Which
propositions are logically equivalent to ?
(implication),
, (contrapositive)
2. Suppose that is known to be
false. Give the truth values for
(a) 
(b) 
(c)
If is known to be false,
then that means when 
is true, 
is false, as
that is the only condition that could cause to be false. For to be true, p has
to be false. For
to be false, q
has to be true. So, we now
know that both p is false and q is true. So:
(a) is the AND of false and true. Thus is false.
(b) is the OR of false and true. Thus is true.
(c) is . We know that is false and p is
false. Thus, is false.
Propositional Equivalences and Implications
3. Show that each of these implications is a tautology by using truth tables
(a) 
(b) 
(c) 
(d) 
(a) 
|
p |
q |
|
|
|
|
q |
|
F |
F |
F |
F |
T |
T |
F |
|
F |
T |
F |
F |
T |
T |
T |
|
T |
F |
T |
F |
F |
T |
F |
|
T |
T |
T |
T |
T |
T |
T |
|
Step |
1 |
2 |
1 |
3 |
1 |
|
All implication values are true. Thus, the implication is a tautology.
(e) (b)
|
p |
q |
r |
|
|
|
|
|
|
F |
F |
F |
T |
T |
T |
T |
T |
|
F |
F |
T |
T |
T |
T |
T |
T |
|
F |
T |
F |
T |
F |
F |
T |
T |
|
F |
T |
T |
T |
F |
F |
T |
T |
|
T |
F |
F |
F |
F |
T |
T |
F |
|
T |
F |
T |
F |
F |
T |
T |
F |
|
T |
T |
F |
T |
F |
F |
T |
F |
|
T |
T |
T |
T |
F |
F |
T |
F |
|
Step |
1 |
2 |
1 |
3 |
1 |
||
All implication values are true. Thus, the implication is a tautology.
(c)
|
p |
q |
r |
|
|
|
|
|
|
F |
F |
F |
T |
T |
T |
T |
T |
|
F |
F |
T |
T |
T |
T |
T |
T |
|
F |
T |
F |
T |
T |
F |
T |
T |
|
F |
T |
T |
T |
T |
T |
T |
T |
|
T |
F |
F |
F |
T |
T |
F |
F |
|
T |
F |
T |
F |
T |
T |
T |
T |
|
T |
T |
F |
T |
T |
F |
T |
F |
|
T |
T |
T |
T |
T |
T |
T |
T |
|
Step |
1 |
3 |
1 |
2 |
1 |
||
All implication values are true. Thus, the implication is a tautology.
(d)
|
p |
q |
|
|
|
|
|
F |
F |
F |
T |
T |
F |
|
F |
T |
F |
T |
F |
F |
|
T |
F |
T |
T |
T |
F |
|
T |
T |
T |
T |
T |
T |
|
Step |
1 |
3 |
2 |
1 |
|
All implication values are true. Thus, the implication is a tautology.
4. 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 |
|
|
|
|
|
|
|
0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
|
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
|
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
|
Step |
2 |
1 |
3 |
1 |
2 |
1 |
|
We show in the above
that 
is a tautology. Therefore, is true. This is actually known as the De Morgan’s
law.
(b)
|
p |
q |
|
|
|
|
|
|
0 |
0 |
1 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
0 |
0 |
1 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
0 |
1 |
1 |
|
Step |
1 |
3 |
1 |
2 |
1 |
|
We show in the above
that 
is a tautology. Therefore, is true. This is actually known as the implication
law.
(c)
|
p |
q |
r |
|
|
|
|
|
|
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
|
0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
|
1 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
|
Step |
1 |
2 |
3 |
2 |
1 |
||
We show in the above
that 
is a tautology. Therefore, is true. This is actually known as the exportation
law.
(d)
|
p |
q |
|
|
|
|
|
|
0 |
0 |
1 |
1 |
1 |
1 |
1 |
|
0 |
1 |
1 |
0 |
1 |
0 |
0 |
|
1 |
0 |
0 |
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
1 |
0 |
1 |
0 |
|
Step |
1 |
3 |
1 |
2 |
1 |
|
When p is
false and q is true, then the above proposition is false, as shown in
the above table. Thus, 
is not a tautology. Therefore, is false.
(e)
|
p |
q |
|
|
|
|
|
0 |
0 |
1 |
1 |
1 |
0 |
|
0 |
1 |
1 |
1 |
1 |
1 |
|
1 |
0 |
0 |
0 |
0 |
0 |
|
1 |
1 |
1 |
0 |
1 |
1 |
|
Step |
3 |
1 |
2 |
1 |
|
When p is
true and q is false, then the above proposition is false, as shown in
the above table. Thus, 
is not a tautology. Therefore, is false.
5. The
“exclusive or” connective 
is defined by the
truth table:
|
p |
q |
|
|
0 |
0 |
0 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |
(a) Show
that 
.
(b) Show that 
.
(a) To show that:
|
p |
q |
|
|
|
|
|
|
|
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
|
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 |
|
|
|
|
|
0 |
0 |
0 |
1 |
0 |
1 |
|
0 |
1 |
1 |
1 |
1 |
0 |
|
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.
6. 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
7. The Sheffer Stroke is a connective 
defined by the truth table:
|
p |
q |
|
|
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 |
|
|
|
|
|
|
0 |
0 |
0 |
1 |
1 |
0 |
1 |
|
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
8. 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, 3; Constructive Dilemmas
5. 1, 4; Transitivity of implication or Hypothetical Syllogism
(b) If 
, 
and 
, then 
.
Proof Explanations
1. Given, or Hypothesis
2. Given, or Hypothesis
3. Given, or Hypothesis
4. 
simplification
5. 1, 4; Hypothetical Syllogism
6. 
5; Rule 25(a)
7. 
3, 7; Modus Ponens
8. 7, 2; Modus Ponens
(c) If 
, 
and 
, then 
.
Proof Explanations
1. Given, or Hypothesis
2. Given, or Hypothesis
3. Given, or Hypothesis
4. 
2; Commutative laws
5. 
4; Implication
6. 
5, 1; Hypothetical Syllogism
7. 
6; Exportation law
8. 
3; Rule 22
9. 
3, 8; Modus Ponens
10. 
9; commutative laws
11. 10, 7; Hypothetical Syllogism
9. Complete the following proof by contradiction 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. 
negation of conclusion
5. 
2, 4; Modus Tollens
6. 
5; De Morgan
7. 6; Simplification
8. 
6; Commutative laws
9. 
8; Simplification
10. 
3;
Commutative laws
11. 
10, 9;
Disjunctive Syllogism
12. 
11, 1;
Modus Ponens
13. 12;
Simplification
14. 
13,
7; conjunction
15. contradiction 14; negation
(b) If , 
and 
, then .
Proof Explanations
1. Given, or Hypothesis
2. Given, or Hypothesis
3. Given, or Hypothesis
4. 
negation of conclusion
5. 4;
De Morgan
6. 4, 1;
Modus Ponens
7. 6;
Simplification
8. 
6;
commutative laws
9. 
8;
Simplification
10. 
9, 2;
Modus Ponens
11. 
7, 10;
conjunction
12. 
11, 3;
conjunction
13. contradiction 12; negation
•
Based on (Ross and Wright 1988) and (Rosen 2003).