Project1e - Rutgers University

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William Yeh
March 10, 2003
Project 1
xxP
( P( x( x) )QxQ
( x())x) Part A:
?<=>?
A-1.
They are equivalent. <=>
A-2.
xxP
( P( x( x) )QxQ
( x())x) We will prove =>
xxP
( P( x( x) )QxQ
( x())x)

(1)
P( x)  Q( x) =>

(3)
P(x) =>
Q (x ) =>
first, using a direct proof.
/
(2) (for an arbitrary xU)
(1) Universal Instantiation
(2) Simplification
(4)
(2) Simplification
x( P( x))
x(Q( x))
=>
(5)
(since x was arbitrary) (3) Universal Generalization
=>
(since x was arbitrary) (4) Universal Generalization
(6)
x( P( x))  x(Q( x)) =>
(7)
(5) + (6) Conjunction
xxP
( P( x( x) )QxQ
( x())x) We will now prove the other direction: => with a direct proof.
xxP
( P( x( x) )QxQ
( x())x)

(1)
/
x( P( x)) =>
(2)
(1) Simplification
x(Q( x)) =>
(3)
(1) Simplification
P(x) =>

Q (x ) =>

(4)
(for an arbitrary xU)
(2) Universal Instantiation
(5)
(for an arbitrary xU)
(3) Universal Instantiation
P( x)  Q( x) =>
x( P( x)  Q( x)) =>
(6)
(4) + (5) Conjunction
(7) (since x was arbitrary for both P(x) and Q(x))
(6) Universal Generalization
Since we have proven both implication directions of the statement, we can conclude that
the statements are indeed equivalent.
A-3.
xxP
( P( x( x) )QxQ
( x())x) Why does imply ?
x( P( x)  Q( x))
The statement , being true, has the meaning, that for every such
object in some universe, each of those objects satisfies two conditions. For example, let
our universe be all students at Rutgers. Let P(x) mean “x is a student” and Q(x) mean “x
attends Rutgers University.” For all x in our universe, that is, students at Rutgers, P(x)
and Q(x) is true, thus is true.
xP( x)  xQ( x)
With this being said, the statement is implied by the first. For
the first statement to be true, every object in some universe must satisfy two conditions.
What this second statement is saying is that for every object x in a universe, some
condition is satisfied, and for every object x in the same universe, some other condition is
also satisfied.
x(Q
P( x))
Returning to our example, we know that every student at Rutgers is
both a student and attends Rutgers University. So we ask ourselves, is every student at
Rutgers a student (that is, )? Yes, of course. Then, is every student at Rutgers attending
Rutgers University (that is, )? Yes again.
In summary, if we know that every object in an universe satisfies two conditions,
we know that every object in that universe satisfies one of the conditions, as well as,
every object in that universe satisfies the second condition.
xxP
( P( x( x) )QxQ
( x())x) Why does imply ?
xP( x)  xQ( x)
The statement has the meaning that for every x in a universe,
some condition is true. Also, for every x in the same universe, some other condition is
true. Knowing that those two conditions are true for all x in an universe, we can
conclude that for all x in a universe, those two conditions are true.
Using the same example mentioned previously, all students at Rutgers, are both,
students, and attend Rutgers. Knowing that all students at Rutgers satisfies those two
conditions, we can say that all students at Rutgers are students and attend Rutgers.
xxP
( P( x( x) )QxQ
( x())x) Part B: ?<=>?
B-1.
The second implies the first. <=
B-2.
xxP
( P( x( x) )QxQ
( x())x) We will prove => using a proof by cases.
xxP
(Q
P( x( x) ))
 xQ( x) We start off with knowing that is true. With this in mind, we know
that one of or MUST be true. And these will be our cases.
x(Q
P( x)) Case 1: is true. We do not care whether is true or not.
xP( x)  xQ( x)
x( P( x))
)  Q( x))

P(x) =>

(1)
(2)
(assumed premise)
(3) (for an arbitrary xU)
/
(2) Universal Instantiation
P( x)  Q( x) =>
(4) (x still arbitrary)
x( P( x)  Q( x)) =>
(3) Addition
(5) (since x was arbitrarily chosen from our universe)
(4) Universal Generalization
x(Q
P( x)) Case 2: is true. We do not care whether is true or not.
xP( x)  xQ( x)
(1)
x(Q
P( x))) Q( x))

Q (x ) =>

(2)
(assumed premise)
(3) (for an arbitrary xU)
/
(2) Universal Instantiation
Q( x)  P( x) =>
(4) (x still arbitrary)
(3) Addition
P( x)  Q( x) <=>
(5)
(4) Commutative Laws
x( P( x)  Q( x)) =>
(6)
(since x was arbitrarily chosen from our universe)
(5) Universal Generalization
xxP
( P( x( x) )QxQ
( x())x) We have shown that in every case, the statement implies .
B-3.
xxP
( P( x( x) )QxQ
( x())x) Why does imply ?
xxP
( P( x( x) )QxQ
( x())x)
The statement, , being true, says that at least one of these
following assertions are true. Either all objects in the universe satisfy some condition or
all objects in the universe satisfy some other condition. The statement says that all
objects in the universe satisfy at least one of two conditions.
xxP
( P( x( x) )QxQ
( x())x)
If the first statement, , is true, we know that all objects in some
universe satisfy some condition. Then, we know for sure that all objects in that same
universe satisfy that condition or some other condition. Therefore, if is true, must be
true as well.
For example, let our universe consist of positive numbers. Let P(x) mean “x is a
positive number” and Q(x) to mean “x is a negative number.” For all positive numbers,
xP( x)  xQ( x) of course, P(x) is true, since all positive numbers are positive numbers.
Therefore the first statement, is true. Let us consider the second statement. Are all
positive numbers either positive or negative? Of course, we already know that all
positive numbers are positive. Therefore, the conclusion is true.
B-4.
xxP
( P( x( x) )QxQ
( x())x) Why is => not true?
xxP
( P( x( x) )QxQ
( x())x)
Here’s a simple counterexample. Let our universe consist of all
people (except hermaphrodites and neuters). Let P(x) stand for “x is male” and Q(x)
represent “x is female.” The first statement asks, are all people male or female? Yes, this
is true. Does this imply the conclusion? The conclusion asks, are all people male or all
people female? This, of course, is not true. Therefore, => is not true, because we have
provided a counterexample which makes the premise true, yet the conclusion is false.
xP
xQ
x) Part C:
x( P( x( x) )Q
( x())
?<=>?
C-1.
The first implies the second. =>
C-2.
xP
xQ
x) We will prove => with a direct proof.
x( P( x( x) )Q
( x())
xP
xQ
x)
x( P( x( x) )Q
( x())

P( s )  Q( s ) =>
P(s ) =>
=>
Q(s ) (4)
(3)
(1)
(2) (for a particular s)
/
(1) Existential Instantiation
(2) Simplification
(2) Simplification
xP(x) =>
(5)
(3) Existential Generalization
xQ(x) =>
(6)
(4) Existential Generalization
xP( x)  xQ( x) =>
(7)
(5) + (6) Conjunction
C-3.
xP
xQ
x) Why is => true?
x( P( x( x) )Q
( x())
x( P( x)  Q( x))
Given the statement being true, we know that there is some x in
our universe that satisfies two conditions. Knowing that, is the conclusion true? The
conclusion states that there is some x that satisfies one condition, P(x), and there is some
x that satisfies another condition, Q(x). From the hypothesis, we already know that there
is some x that satisfies both P(x) and Q(x). Therefore, that x satisfies P(x) and that x
satisfies Q(x) and the conclusion is also true.
For example, let our universe be people and let P(x) stand for “x is named
Stephen Max” and let Q(x) represent “x teaches cs205.” The hypothesis is true, so we
know that there is someone named Stephen Max that teaches cs205. Then, the
conclusion asks, is there a person named Stephen Max and is there a person that teaches
cs205? The answer is yes, there is someone named Stephen Max, and there is a person
that teaches cs205, namely, Stephen Max.
C-4.
xP
xQ
x) Why is => not true?
x( P( x( x) )Q
( x())
Here’s a simple counterexample. Let our universe be people, and let P(x) stand
for “x is named Stephen Max and x teaches cs205” and let Q(x) represent “x is named
William Yeh and x takes cs205.” There does exist someone named Stephen Max who
teaches cs205. There also exists someone named William Yeh who is taking cs205.
Then is the conclusion true? Is there someone named Stephen Max and William
Yeh and who teaches cs205 and takes cs205? Maybe, but we cannot know for sure from
our given hypothesis. Therefore the conclusion can be either false or true given that the
hypothesis is true, which in logic, makes the entire implication false.
xP
xQ
x) Part D:
x( P( x( x) )Q
( x())
?<=>?
D-1.
They are equivalent. <=>
D-2.
xP
xQ
x) We will prove that => with a proof by cases.
x( P( x( x) )Q
( x())
xP
xQ
x)
x( P( x( x) )Q
( x())

P( s )  Q( s ) =>
(1)
(2) (for a specific s)
/
(1) Existential Instantiation
P( s)  Q( s) Here is where the proof by cases comes in. If is true, then either P(s) or
Q(s) MUST be true, and these will be our cases.
Q
P(s ) Case 1: is true. And we do not care if is true.
P(s )
xP(x)
=>
(3) (for a specific s)
Assumed Premise
(4)
(3) Existential Generalization
xP( x)  xQ( x) =>
(5)
(4) Addition
Q
P(s ) Case 2: is true. And we do not care if is true.
Q(s )
xQ(x) =>
(3) (for a specific s)
(4)
Assumed Premise
(3) Existential Generalization
xQ( x)  xP( x) =>
(5)
(4) Addition
xP( x)  xQ( x) <=>
(6)
(5) Commutative Laws
xP
xQ
x) We have proven that => for all cases.
x( P( x( x) )Q
( x())
xP
xQ
x) We will prove that => also with a proof by cases.
x( P( x( x) )Q
( x())
xQ((x
(x
xP
x))  xQ( x) We know that is true, and therefore either or MUST be true. These
will be our cases.
xQ(x
(x)) Case 1: is true. And we do not care if is true or false.
xP
xP( x)  xQ( x)
(1)
xP
x( P(x( x) )  Q( x))

(2)
(assumed premise)
/
P(s )
P( s )  Q( s ) =>
=>
(3) (for a specific s)
(4) (for same specific s)
x( P( x)  Q( x)) =>
(2) Existential Instantiation
(3) Addition
(5)
(4) Existential Generalization
xQ(x
(x)) Case 2: is true. And we do not care if is true or false.
xP
xP( x)  xQ( x)
(1)
xQ
x( P(x
( x))  Q( x))

(2)
Q(s ) =>
(assumed premise)
(3) (for a specific s)
/
(2) Existential Instantiation
Q( s )  P( s ) =>
(4) (for same specific s)
(3) Addition
P( s )  Q( s ) <=>
(5)
(4) Commutative Laws
x( P( x)  Q( x)) =>
(6)
(5) Existential Generalization
xP
xQ
x) We have proven that => is true for all cases.
x( P( x( x) )Q
( x())
xP
xQ
x) We have proven both directions of the statement <=> . Therefore,
x( P( x( x) )Q
( x())
they are equivalent.
D-3.
xP
xQ
x) Why is => true?
x( P( x( x) )Q
( x())
The hypothesis states that there is some x in our universe that satisfies at least one
of two conditions. The conclusion states that either one of the following is true:
something satisfies condition one or something satisfies condition two. Given that there
exists something that satisfies one of two conditions (the hypothesis), we know for sure
that there exists something that satisfies condition one or there exists something that
satisfies condition two.
For an example, let our universe be people, let P(x) stand for “x plays for the
Houston Rockets” and Q(x) represent “x plays for the Los Angeles Lakers.” We know
that there exists someone that either plays for the Rockets or the Lakers. Let our x be
Yao Ming. Does Yao Ming play for the Rockets or the Lakers? Yes, so the hypothesis is
satisfied. We turn to the conclusion. Our x is still Yao Ming, so we ask ourselves, does
Yao Ming play for the Rockets, or does Yao Ming play for the Lakers? Yao Ming plays
for the Rockets, so the conclusion is also satisfied.
Let’s just take another example to further cement this implication in our minds.
Using the same example, let’s find another x to satisfy the hypothesis. Shaquille O’Neal
plays for the Lakers. So our hypothesis is satisfied. So is there someone who plays for
the Rockets? We don’t know. Is there someone who plays for the Lakers? Yes, we
know that Shaquille O’Neal plays for the Lakers. Therefore, the conclusion follows from
the hypothesis again.
xP
xQ
x) Why is => true?
x( P( x( x) )Q
( x())
The hypothesis states that there exists something in our universe that satisfies a
condition, or there exists something in our universe that satisfies some other condition.
Given that we know that something does indeed satisfy either the first condition or
something exists that satisfies the second condition, it follows that there exists something
that satisfies either the first or the second condition.
For example, let our universe be people, and let P(x) stand for “x is almost done
with Project 1” and let Q(x) stand for “x has not even started Project 1.” So, is there
someone who is almost done with Project 1? Yes, his name is William Yeh. So the
hypothesis is true. Given that we know that William Yeh is almost done with Project 1,
does the conclusion hold? The conclusion asks, is there someone who is almost done
with Project 1 or has not even started Project 1? From the hypothesis, we know that
William Yeh is almost done with Project 1, and therefore the conclusion is true, and thus
the conclusion follows from the hypothesis.
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