0. Logical Foundations
1 Proving Mathematical Statements
In this introduction you will study the basic components of
proofs
.
How to prove theorem is at
the heart of mathematics. Our goal is to learn the process of proof. Along the way we will
en
counter some of the fundamental concepts of mathematics. This could serve as a nice prelude to
Abstract Algebra.
Mathematicians often refer to proven results as Theorem, Proposition, Lemma or Corollary.
They use “
Theorem
”
for the most important res
ults. A
Corollary
is a result that follows easily
from the result(s) preceding it; sometimes corollaries are important results themselves.
A
Lemma
is a result that is not of great interest in itself, but is used as an important tool in
proving forthcoming
propositions or theorems. In these notes, with a couple of exceptions, results
are called “Theorems”.
It will be your job to provide proofs of all theorems, corollaries, etc., that are not already
proved in these notes, and to solve all of the exer
cises.
Most theorems are statement of the form
“ If, P then Q ”,
(
symbolically as
“P
Q
”
)
where
P
and
Q
are statements themselves. In this case
P
is called the
hypothesis
and
Q
is called
the
conclusion
.
A
statement
( or a
proposition
) we refer to a sentence that is either
true
or
false
, but not both.
(Technically, a statement involving a variable is called an
open sentence
.)
Let us consider some examples. We will analyze and discuss strategies for proving
theorems in
general and these examples in particular. Let
A
be the statement
“the square of any
even integer is even”
and let
B
be the statement
“if the square of an integer is even, then the
integer is even”
. Both A and B are true statements, but they are proved in
rather different ways.
Before starting, we should note that in almost all mathematical theorems, the variables
are only allowed to take on certain values, that is, to come from certain classes of objects.
For example, in the statement “ x·y = y ·x ”,
we generally assume x, y are numbers. Indeed,
the statement does not make sense if x, y are people and the statement is false if x, y are
allowed to be matrices.
In the statements A and B above, all the objects considered are supposed to be integers,
that is, positive and negative natural numbers. All variables represent numbers from the infinite
list
. . .,

3,

2,

1, 0, 1, 2, 3, . . . .
We denote the set of all integers by
Z
.
Let’s start by identifying the hypothesis and the conclusion in ea
ch of A and B. Let
x represent an integer, let P be the open sentence “
x is even
”, and let Q be the open
sentence “
x
2
is even
”. Then B is simply the statement “Q
P ”, and so the hypothesis
of B is
“x
2
is even
”, while the conclusion is
“x is even”.
At fir
st, A may not appear to
be an “if
–
then” statement, but in fact it is. A can be re

phrased as follows:
“for every integer x, if x is even, then x
2
is even”,
which we write symbolically
(
x
Z)(P(x)
Q(x)) .
Thus the
hypothesis of A is
“x is even”
and the conclusion is “
x
2
is even
”.
Similarly, the statement B should be understood as the statement
(
x
Z)(Q(x)
P(x)) .
Once we have proved both, we will have proved
(
x
Z)(Q(x)
P(x))
In English, we say “
P(x) is true if and only if Q(x)
is true”, or
“P(x) is equivalent to Q(x)”.
There are usually several ways to prove a theorem. We now describe a few general strategies.
Direct Proof
.
The assertion
“P
Q”
means that if
P
is
true, then
Q
must be true.
To prove
P
Q
directly, we assume that P is true and show by a sequence of logically
connected arguments that the truth of Q follows.
This seems to be the easiest way to prove statement A. The hypothesis that x is even can be
re

written as
“x = 2n for some integer n”.
Thus
x
2
= 4n
2
= 2(2n
2
).
If we set
m = 2n
2
, then m is an
integer and
x
2
= 2m
. This proves x
2
is even.
Contrapositive Proof.
Given a statement S, the negation
(not S)
of S is a statement
that is true precisely w
hen S is false. For example
“ not (Today is Sunday)”
is
“Today is not Sunday”.
The statement
“P
4´
means that the truth of P implies the truth of Q. This is logically
equivalent to the statement that Q is false implies P is false. Thus
“P
4´
is logi
cally equivalent
to
“( not Q )
(not P) ”.
This latter statement is called the
contrapositive
of the implication
“P
Q”.
To prove
P
Q
by a
contrapositive
argument, we prove the logically equivalent statement
“
not Q
n潴⁐
”. That is, we show tha
t if Q is assumed to be false (its negation not Q is
assumed to be true), then P must be false (its negation not P is true).
If we try to prove A this way, we see that we should assume x
2
is not even, and we must prove
x is not even. Unfortunately, it
is not obvious what we should do next. (Well, “not even” is
equivalent to “odd”, so the contrapositive is equivalent to “if x
2
is odd, then x is odd”. It’s not
clear where to go from there, however.)
Statement B, however, fares a bit better with a cont
rapositive proof. Recall that B is the
statement Q
P , so its contrapositive is the statement “ not P
not Q”. Thus we may assume
not P , i.e., we may assume x is not even. But this means x is odd. We now must show not Q, that
is, we must show x
2
is odd
.
Exercise 0.1.
Complete the proof of statement B, using the contrapositive approach
outlined above.
Proof by Contradiction.
To prove a statement S by contradiction, we assume the statement S is
false. We then show that this assumptio
n leads us to a statement which is known to be incorrect
(possibly a contradiction to a previously proven theorem or, if we’re lucky, an obvious absurdity
like 0 = 1). It follows that our assumption of the falsity of S must have been wrong. Hence S must
be
true.
To prove
P
Q
by contradiction, we assume that
P
Q
is false. The only way this can
happen is if P is true and Q is false. Thus we may assume P is true and Q is false, and we must
obtain a contradiction.
For example, to prove B (which is
“Q
P
”) by contradiction, we would assume P is false and
Q is true. That is, we would assume x is odd (not even) and x
2
is even. But we saw in the
contrapositive proof (did we?) that x is odd implies x
2
is odd, so we now
have that x
2
is both even
and od
d. This is a contradiction
—
no integer is both even and odd
—
and so we have proved B.
Exercise 0.2
. If you haven’t already, prove that the square of an odd integer is odd.
Exercise 0.3
.
Prove statement A (if x is even, then x
2
is even) by contradictio
n. You
will probably want to follow the pattern of the proof of B by contradiction.
Exercise 0.4
. Prove that if x is an integer and x
2
is odd, then x is odd. Use any method you wish.
In the course of doing mathematics, it often happe
ns that disproving a statement provides valuable
information. Since a statement is considered mathematically true only if it holds in general (i.e.,
all of the time), the following is a valuable method for disproving statements.
Counterexample.
To prove a
statement is false it suffices to exhibit one example for which the
statement does not hold.
Warning.
To show that a statement is true, it is not sufficient to present an example for which the
statement holds!
Consider the statements C:
“the sum of two od
d integers is odd
” and D:
“the square of any integer
is positive”.
Statement C is false, and we would typically show this in the following way: 1 and 3
are odd numbers, but their sum 1 + 3 = 4 is not odd. Thus statement C is false. Statement D is also
fals
e as the counterexample 0
2
= 0 shows. Presenting any number of positive examples, 1
2
= 1 > 0,
2
2
= 4 > 0, (

3)
2
= 9 > 0, does not show that statement D is true.
Finally, consider the statement E:
“any even integer greater than 2 is the
sum of two prime numbers”
.
You can give many examples where E is true, e.g., 4 = 2+2, 6 = 3+3, 8 = 5 + 3, 10 = 7 + 3
(or 10 = 5 + 5), 100 = 53 + 47, 1000 = 653 + 347, and so on. In fact, no one knows of a
counterexample. Nonetheless, statement E has not
been proven, and its truth is still an open
question.
The guess that E is a true statement is known as
“Goldbach’s Conjecture”.
Exercise 0.5
. If possible, identify the hypothesis, the conclusion, and any variables (explicit or
implicit) in the following statements.
(a) The product of two odd numbers is odd.
(b) The sum of two odd numbers is odd.
(c) There is an odd number whose square is 100.
Exercise
0.6
. Prove or disprove the statements in Exercise 0.5.
Not all statements we wish to prove have the from “P
Q”, of course. For example, Exercise
0.5(c) does not appear to be of this form. Another example is given by the theorem
“x
3
+ y
3
= (x + y)(x
2

xy + y
2
) for any numbers x, y
”.
Most theorems do come down to P
儠獴Q瑥浥湴猬琠nea獴渠摩獧畩獥⸠u潲xa浰me,
瑯⁰牯癥⁴桡琠堠
Y = Y
X for sets X, Y, we actually need to verify that an element z
is in the set X
Y if and only
if z is in the set Y
X. Symbolically, we need to show
“ z
X
Y
z
Y
X ”, and this is equivalent to the statement (z
X
Y
z
Y
X) and
(z
Y
X
z
X
Y).
Thus to prove X
Y = Y
X, we actually need to prove two “P
Q”

type statements,
namely “
z
X
Y
z
Y
X ” and “ z
Y
X
z
X
Y ”
Exercise 0.7.
Prove x
3

y
3
= (x

y)(x
2
+ xy + y
2
) for all numbers x, y.
Exercise 0.8.
Prove A
B = B
A for all sets A,B.
Some notation:
Our main goal is to learn how to understand and communicate deductive
re
asoning. We will still be doing mathematics, however, and we will use some familiar examples,
many of them involving numbers. We now introduce some common types of numbers and the
notation mathematicians use for them. In doing so, we will use the set notat
ion to be discussed in
Section 2.
The
natural numbers
are the counting numbers including 0, and this set is denoted by N. That is,
N = { 0, 1, 2, . . . }. Not every mathematician agrees with this terminology. Some people specify
that the natural numbers s
tart at 1 instead of 0. We will instead refer to the set of numbers
{ 1, 2, . . . } as the positive integers; for us the natural numbers are the non

negative integers.
The set of integers is made up of the natural numbers together with their negatives (ad
ditive
inverses). This set is denoted by Z, and so Z = { . . . ,

2,

1, 0, 1, 2, . . . }. The rational numbers are
all of the usual fractions; the set is denoted by Q. Thus
Q = {
a
b
 a, b, are integers and b
0 }.
The
set of real numbers is denoted R. This set consists of all the numbers on the number line,
including both the rational numbers and irrational numbers like
2
or
.
Real numbers make up all the numbers that have a decimal expansion (even one th
at never ends
or has no pattern).
Mathematics has been in use for thousands of years, but a satisfactory axiomatic, formal
construction of the number system was only begun in the 19th century and was not fully achieved
until the 20th century. And not long
after it was fully achieved, it was shown by Kurt G¨odel
that a perfect axiomatic construction is impossible! G¨odel’s famous result is beyond the
scope of this class; we will not even state it. One thing I hope you will take from this is that while
mathe
matics has a very elegant and condensed structure that is beautiful but often seems daunting
or even inhuman, that structure is in fact the work of many people over hundreds of years, and
there remain many questions to be answered and improvements to be ma
de.
1.
Logic and the Language of Mathematics
The language and rules of reasoning used in mathematics are mostly the same as those used in
everyday life. Nevertheless, we need to discuss mathematical language and some basic notions of
logic. We hav
e tried to include just enough discussion to get our mathematical studies off to a
smooth start.
One problem with everyday English is that words permit different interpretations, sometimes
leading to misunderstandings. In mathematics this is unacceptab
le. We never want a mathematical
sentence to have ambiguous meaning. For this reason we will discuss a few key words and
phrases and their precise meanings when used in mathematics. In the process we will introduce
some useful notation.
A
statement
is
a sentence which is meaningful and is either true or false. When we speak of the
meaning of a statement, we refer to its truth value: “true” or “false”. An open sentence is a
sentence involving some unknowns (variables) that becomes a statement when a leg
itimate value
is assigned to each of the variables. For example, “
For any number x, x
2
= (

x)
2
”, “
3 + 4 = 9
”,
“
There are sets A, B with A
B = A
B
”, “
All functions are one

to

one
”, and
“ The Moon is
larger than Pluto
” are all statements (not necess
arily true statements).
The sentences “
x
2
+ 5 = 4x + 1
”, “ A
B = A”, and “ f is one

to

one” are open sentences: these
sentences become true or false when we specify values for variables x, A,B, f . It is understood
here that x must a number, A, B mu
st be sets, and f must be a function. The sentence “A
1 = 1”
is not an open sentence since we cannot take the union (
) of a set A and the number 1; thus even
if we replace A by a specific set, the sentence does not become true or false: it is not meaning
ful.
Similarly, “
A or B = B or A
” is not an open sentence: we cannot combine sets with the
or operator. On the other hand, “
A
{1} = {1}”
and “
x
A or x
5
” are open sentences.
Technically, the sentence “
x
2
= (

x)
2
” is an open sentence. However, it
is true no matter what
number we substitute for x, so we often think of “x
2
= (

x)
2
” as a true statement, and many
mathematical theorems are stated this way. The actual statement or theorem, though, is the
sentence “
for all numbers x, x
2
= (

x)
2
”. We wil
l discuss this later in this section when we talk
about quantifiers.
Compound sentences
Many mathematical statements and open sentences are compound, that is, they are the
combinations of several simpler statements or open sentences. Examples of this are
the statements
“Given any integers x, y, if x < y, then x < y

1 or x = y

1” and “ If x is an element of A
B,
then x
A and x
B”. We need to briefly discuss the ways in which statements and open
sentences are combined. Let P and Q be statements
or open sentences. We will refer to them as
statements below. If they are open sentences, then “true” and “false” should be interpreted as
meaning “become true [or false] when variables are given legal values”.
“P and Q”
is a statement which is true w
hen statements P and Q are both true and is false
otherwise. This is also written “P
Q”.
“P or Q
” is a statement which is true when either statement P or statement Q (or both P and Q)
is true and is false when both P and Q are false. This is also wr
itten“P
Q”.
For example, the statements “3 + 4 = 7 and 3 · 4 = 12” and “3 + 4 = 7 or 3 · 4 = 12” are both
true. (The second statement is true because “or” for us is inclusive or.) However, the statement
“3 + 4 = 7 and 3 · 3 = 12” is false, while the st
atement “3 + 4 = 7 or 3 · 3 = 12’ is true.
Notice the above definitions are symmetric in P and Q, i.e., the statement “Q and P ” has the same
meaning as “P and Q”, and “Q or P ” has the same meaning as “P or Q”.
“ P implies Q
”, or equivalently “ if
P then Q ”, is a statement which is true unless P is true and
Q is false. Hence P implies Q is true if either P is false or Q is true. “P implies Q” is often
abbreviated as “P
Q”.
The statement “Q
P ” is called the
converse
of “P
Q”. These two stat
ements do not have the
same meanings: you are asked to give an example of this in Exercise 1.2.
The meanings of the words and, or, and implies are often presented in “truth tables”. Truth tables
can be a useful device, but we will not require them in this
class. There is a brief discussion of
truth tables and a presentation of the various logical operators (and, implies, etc.) in terms of truth
tables at the end of this section.
“P if and only if Q”
( abbreviated “P
Q”) is true if either both P and Q
are true, or both P and
Q are false. Thus P
Q is true when P and Q have the same truth value. When P
Q is true we
say the statements P and Q are
equivalent
.
Note that “P
Q” is equivalent to “ P
Q and Q
P ”. When we wish to prove a statement
o
f the form “P
Q”, we frequently break it up into two parts and prove that both “P
Q” and
“Q
P ” are true.
“not P ”
is a statement, called the negation of P, which is true if P is false and is false if P is
true. It is sometimes written “¬P ” .
T
he statement ( not Q
not P) is called the
contrapositive
of the statement P
Q. Notice that
( not Q
not P) is true unless P is true and Q is false. Hence the statement P
Q is equivalent
to its contrapositive. This observation is quite useful in doi
ng proofs.
Exercise 1.1
.
The following sentences may be viewed as “P ) Q” statements. In each case,
identify the statements P and Q, then write (in words) the converse and
contrapositive of the sente
nces.
(a) The sentence: “ If it is raining then there are clouds in the sky.”
(b) The sentence: “ If x
2
= x, then x = 1.”
Exercise 1.2
. Construct statements P , Q, and R (different from those in Exercise 1
.1) such that
(a) P
Q is true but the converse is false.
(b) Q
R is true.
(c) What can be said about the statement P
R?
Exercise 1.3.
Explain why the statement “ not (
P and Q)” is equivalent to the statement
“( not P) or ( not Q)”.
Exercise 1.4.
In the spirit of Exercise 1.3, find a statement equivalent to “ not (P or Q)”,
and explain your answer.
Exercise 1.5
. Explain why the
statement “P
Q” is equivalent to the statement
“ not P
not Q”.
Exercise 1.6
.
Explain why the statement “(P and Q) and R” is equivalent to the statement
“P and (Q and R)”.
Sample Solution:
The statement “
(P a
nd Q) and R
” is true precisely when both the statement “
P and Q
” and the
statement
R
are true. “
P and Q
” is true precisely when both
P
and
Q
are true. Thus “(
P and Q) and R
” is true
precisely when all 3 of P, Q, R are true.
Similarly “
P and (Q and R)”
is true precisely when P and
“Q and R”
are both true, and this happens precisely
when all of P, Q, R are true. Thus the two statements
“(P and
Q) and R” and “
P and (Q and R)”
are true for exactly
the same truth values of the statements
P, Q, R
(the variab
les). This proves they are equivalent.
Exercise 1.7.
Explain why the statement “(P or Q) or R” is equivalent to the statement
“P or (Q or R)”.
Exercise 1.8
.
Explain why the statement “P and (Q or R)” is equivalent to the statemen
t
“(P and Q) or (P and R)”.
Exercise 1.9.
In the spirit of Exercise 1.8, find a statement equivalent to “‘P or (Q and R)”,
and explain your answer
.
Exercise 1.10
. Explain why if [(P
Q) and (Q
R)] is true, the
n (P
R) must be true.
Terminology 1.A
. A statement that is true based solely on its logical form (equivalently, one
that can be verified by a truth table) is called a
tautology
. We will discover other tautologies as
necessary.
Next we conside
r
open sentences
—
these are sentences which contain one or more variables
and which become statements when the variables are replaced with actual objects. For example
“x > 5” becomes a true statement when x is replaced with “7”. It becomes a false statem
ent when
x is replaced by “3”. If P(x) represents our open sentence, then P(7) is the true statement “7 > 5”
and P(3) is the false statement “3 > 5”.
Open sentences can be joined by the logical connectives and () , or ,
,
and be modified by
not .
The meaning of these compound open sentences is the meaning of the
corresponding compound statements when the variables are replaced by actual objects.
Quantifiers
Another way to turn open sentences into statements is with quantifiers.
The
universal
qua
ntifier is the phrase “
for every
” (or equivalently “
for all
”) and is denoted
“
´
.
The
existential
quantifier is the phrase “
there exists
” (or equivalently, “for some”), is denoted
“
´
If P(x) is an open sentence, then
“(
x)(P(x))”
—
read “
for all x, P
(x)”
—
states
that P(x) is true for all possible values of x. The expression
“(
x)(P(x))”
—
read “
there
exists an x such that P(x)”
—
states that there is at least one value of x for which P(x)
becomes a true statement.
If P(x) is the open sentence “x
> 5”, then
“(
x)(P(x))”
is the statement “
for all objects x, x > 5
”.
The expression
“(
x)(P(x))”
is the statement “
there exists an object x such that x > 5
”.
When quantifiers are used, it is usually understood or explicitly stated that the variable(s) com
e
from a particular class. In the example above, this class might be the class of positive integers or
the class of real numbers. We sometimes make this explicit by writing (
x
X)(P(x)), where X is
the class in question. See next Section for more on sets
and the symbolic notations.
Thus if we understand that x represents a non

negative integer, then “(
x)(x >

1)” is true.
However, if x represents a real number, the statement is false. That is, “(
x
N)(x >

1)” is true
and “(
x
R)(x >

1)” is false
.
Exercise 1.11
. Write an open sentence with one variable x. Letting Q(x) denote your open
sentence, write English sentences for the statements “(
x)(Q(x))” and
“(
x)(Q(x))”.
Exercise 1.12
.
Let
P(x) be an open sentence in the variable x. If “(
x)(P(x))” is true, must
“ (
x)(P(x))” also be true? What if we relativize the statements to a set X: does
“(
x
X)(P(x))” imply “(
x
X)(P(x))”?
T
here is a surprising connection between the quantifiers
and
. The statement
“
not [(
x)(P(x))]”
is the statement “
not (for all objects x, P(x) is true)”
or equivalently “
P(x) is
false for at least one object x
”. This last formulation is clearly equiva
lent to
“(
x)(P(x) is false
)”
or simply
“(
x)( not P(x))”.
That is,
“ not [(
x)(P(x))]”
is equivalent to
“ (
x)( not P(x))”.
Exercise 1.13
. Use a line of reasoning similar to the above paragraph to show that the statement
“ not [(9x)(P(x))]” is equivalent to “(8x)( not P(x))”.
Exercise 1.14.
Let P(x) be the statement “x > 5”.
(a) Write out sentences for “ not [(
x)(P(x))]” and “(
x)( not P(x))”.
(b)
Do the same for the statements “ not [(
x)(P(x))]” and “(
x)( not P(x))”.
When a variable falls under the scope of a quantifier in a statement, it is no longer a free
variable: we cannot substitute values for it without changing the meaning of
the sentence, or
making it meaningless. Such a variable is also a dummy variable in the sense that we can replace
it by any variable not occurring in the sentence without changing the meaning.
For example, the sentence “x = 0 and (
z)(
y)(z
y)” is de
noted by P(x) and not P(x, y, z),
since y and z cannot be given values without changing the meaning or making the sentence
meaningless. However, this sentence is equivalent to the sentence “x = 0 and ((
x)(
z)(x
z))”.
(The second and third occurrences of
x, and both occurrences of z, are “dummy” occurrences
because of the quantifiers
and
.) If we replace z by x in this new sentence, however, we do
change the meaning.
To sum up:
Strictly speaking, an open sentence
—
that is, one with free occurrence
s of
variables
—
is neither true nor false. When we say a statement like x + y = y + x is true, we mean
that it becomes a true statement when a universal quantifier
is inserted in front for every free
variable. Thus when we say x + y = y + x is true, we
mean the statement (
x)(
y)(x + y = y + x)
is true. When we say the statement x

y = y

x is not true, we really mean that the statement
(
x)(
y)(x

y = y

x) is not true.
Equality
The last basic concept we discuss in this section is equality. Wh
en two objects x and y are
identical, we say they are equal and we write x = y. We will not give a formal definition of
equality in general; rather when we introduce a concept (e.g., set, function, etc.), we will say what
it means for two objects of that t
ype to be equal.
There are two main properties of equality, which we state as axioms. The first states that
anything is equal to itself; the second says roughly that if two objects are equal, the second may be
substituted for the first in any sentence
without changing the truth value of that sentence.
Axiom 1.B(Equalaty)
.
x = x is true for any object x
.
Axiom 1.C (Substitution Principle for Equality).
Let x, y be variables, let P be a sentence
containing the variable x, and suppose
that x and y are free variables in P . (That
is, x, y do not come under the scope of any quantifiers.) Let P
’
be a sentence
obtained from P by substituting y for some (or all) of the occurrences of x.
Then (
x = y)
⡐
P
0
⤠楳⁴)ue瑡 ement
.
Example 1.D
. The formula (x + z)
2
= x
2
+ 2xz + z
2
is called the Binomial Theorem; it is true for
any numbers x, z. Suppose x = y. Then the Substitution Principle for Equality implies that
(x + z)
2
= y
2
+ 2xz + z
2
is
true. We can sum this up with either of the following two sentences.
The first is more “English

friendly”, while the second is more formal. (More formal is not always
better.)
If x = y, then (x + z)
2
= y
2
+ 2xz + z
2
.
or
(
x)(
y)(
z) ((x = y)
( (x + z)
2
= y
2
+ 2xz + z
2
)_.
Exercise 1.15
.
Give an example of the Substitution Principle for Equality in an English sentence.
Two other extremely important proper
ties of equality are listed in the next theorem. They can be
proved from Axioms 1.B and 1.C.
Theorem 1.16.
For any objects x, y, z, the following statements are true.
(a) (Symmetry) If x = y, then y = x.
(b) (Transitivit
y) If x = y and y = z, then x = z.
Proof.
(a) Suppose x = y. By Axiom 1.B, we know x = x. Now by Axiom 1.C, we can
replace the first x in this last statement by y, and we obtain y = x.
(b) Proving this part is your job
.
Exercise 1.17
.
Use the Substitution Principle for Equality to prove x = y
x+z = y+z.
Some more abbreviations:
“iff”
is an abbreviation for
“if and only if”.
“
” and “&” are sometimes
used as abbreviations for the logical connective “and”, while “
”
is often used as an abbreviation
for “or”. “¬” and “~” are sometimes used as abbreviations for “not”.
“
´
is an abbreviation for “there exists a unique” or equivalently, “there exists exactly one”.
“s.t.”
and
“s.th.”
are abbreviations for “such that”.
Truth Tables
The meanings of logical connectives, and the proofs of logical statements, are often presented by
“
truth tables”.
These are tables that list all possible combinations of truth values of the variables,
and that show the truth values of th
e statements we are interested in.
For example,
and
,
or
, and
implies
are often presented in truth tables as follows.
P
Q
P
Q
T
T
†
T
T
F
†
F
F
T
†
F
F
F
†
F
P
Q
P
Q
T
T
T
T
F
F
F
T
T
F
F
T
P
Q
P
Q
T
T
†
T
T
F
†
F
F
T
†
F
F
F
†
F
“P if and only if Q” (“P
Q”) is true if either both P and Q are true, or both P and Q are false.
The truth table is below. Recall that when P
Q is true we say the statements P and Q are
equivalent
,
not equal.
P
Q
P
Q
T
T
T
T
F
F
F
T
F
F
F
T
倠
Q is equivalent to (P
Q and Q
P ). This is obvious from the definition, but it may also
be seen by comparing the last column of the truth table given above with the last column of the
one compiled below.
P
Q
P
Q
Q
P
⡐
儩
⡑
倩
T
T
T
T
T
T
F
F
T
F
F
T
T
F
F
F
F
T
T
T
“not P ” (“ ¬P ”) the negation of P , which is true if P is false and is false if P is true, has the
following truth table.
P
not P
T
F
F
T
The
contrapositive
of P
Q is the statement ( not Q
not P). We’ve seen befor
e that P
Q is
equivalent to its
contrapositive
. We can also prove this by noting that the last two columns in the
following truth table are the identical.
P
Q
not P
not Q
(not P)
⡮潴⁑)
倠
Q
T
T
F
F
††††††
T
T
T
F
F
T
††††††
T
T
F
T
T
F
††††††
F
F
F
F
T
T
††††††
T
T
Exercise 1.18
. Show by constructing a truth table that the statement P
Q is not equivalent to its
converse Q
P .
Exercise 1.19.
Show by constructing truth table(s) that the s
tatement
“ not (P and Q) ”
is
equivalent to the statement “
( not P) or ( not Q)
”.
Hint:
To save time, combine all your work into a single table.
Exercise 1.20.
In the spirit of Exercise 1.19, find a statement equivalent to
“ not (P or Q)”. Use a
truth table to verify the equivalence.
Exercise 1.21
. Find a statement equivalent to “ not (P
Q)”. Verify with a truth table.
The following exercises all involve “compound statements” based upon three ini
tial statements.
They require a larger truth table.
Exercise 1.22.
Show by constructing truth table(s) that the statement “
(P or Q) or R
” is
equivalent to the statement “
P or (Q or R)
”.
Hint:
You need to construct a truth
table (or truth tables) with eight rows.
Note:
A priori the statement “P or Q or R” does not make sense, but Exercise 1.22 shows that
we can take it to mean either “(P or Q) or R” or the equivalent “P or (Q or R)”.
Exercise 1.23.
Show by
constructing a truth table that the statement “
P and (Q or R)
”
is equivalent to the statement “
(P and Q) or (P and R)
”.
Exercise 1.24
. Show by constructing a truth table that the statement “
P
(Q or R)
”
is equivalent to the statement “
[P and ( not Q)]
R
”.
Exercise 1.25.
Prove that if [(P
Q) and (Q
R)] then (P
R), both with a truth
table and by reasoning without using a truth table.
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