Michael Tsyvine
Section: 8
Discrete Project 1
Introduction: In this project, I will verify which directions of certain implications and
propositions in predicate logic hold true for all cases and which do not. I will do this by
employing formal proofs, informal explanations of the proofs, and counterexamples
where necessary. By doing this, I hope to establish additional rules that can be used
freely and correctly in future proofs that are more complicated that the ones used to prove
the rules themselves.
Let P(x) represent “x is a painter” and Q(x) represent “x is a cook” for the informal
explanations.
PART A
x(P(x)  Q(x))  xP(x)  xQ(x)
This implication is a bi-conditional. It holds true in both directions. To prove this, I will
use a direct proof.
Proving this direction: 

1. P(x)  Q(x)

2. P(x)

3. xP(x)
Universal Instantiation (since
for all x in the universe of
discourse, P(x) and Q(x) are true,
we can say that for an arbitrary x,
P(x) and Q(x) will also be true.
These free variables are derived
from universally quantified
expressions so we know that they
will hold true.)
1. Simplification (since we
know
that P(x) and Q(x) are true, P(x)
alone must be true.)
Universal Generalization
(since an arbitrary x was used to instantiate P(x), we can now make the general statement that for
all x, P(x) will be true. As I said before, all free variables are derived from universal quantifiers, and not
existential ones.)

4. Q(x)
1. Simplification (since we
know
that P(x) and Q(x) are true, Q(x)
alone must be true.)

5. xQ(x)
Universal Generalization
(since an arbitrary x was used to instantiate Q(x), we can now make the general statement that for
all x, Q(x) will be true. As I said before, all free variables are derived from universal quantifiers, and not
existential ones.)

6. xP(x)  xQ(x)
3,5 Conjunction since we know
that both expressions are true, the
conjunction of the two expressions
must hold true by definition of
conjunction.
Q.E.D.
1. x(P(x)  Q(x))  xP(x)  xQ(x)
Proving this direction: 

2. xP(x)
1. Simplification (since we

3. xQ(x)
know
that xP(x) and xQ(x) are true,
xP(x) alone must be true.)
1. Simplification (since we
know
that xP(x) and xQ(x) are true,
xQ(x) alone must be true.)

4. P(x)
2. Universal Instantiation
(since for all x in the universe of
discourse, P(x) is true, it must be
true when we choose an arbitrary x
value.)

5. Q(x)
3. Universal Instantiation

6. P(x)  Q(x)

7. x(P(x)  Q(x))
(since for all x in the universe of
discourse, Q(x) is true, it must be
true when we choose an arbitrary x
value.)
4,5 Conjunction since we know
that both expressions are true, the
conjunction of the two expressions
must hold true by definition of
conjunction.
Universal Generalization
(since an arbitrary x was used to instantiate both P(x) and Q(x), we can now make the general
statement that for all x, P(x) and Q(x) will be true. As I said before, all free variables are derived from
universal quantifiers, and not existential ones.)
Q.E.D.
Informal Explanation: This is an equivalence because if it is the case that everyone is a
painter and a cook, then it is logically the case that everyone is a painter and everyone is
a cook. Similarly, if everyone you ask is a painter and everyone you ask is a cook
(including those that are already painters), then everyone is obviously going to be a
painter and a cook.
PART B
x(P(x)  Q(x))  xP(x)  xQ(x)
This implication only holds true in the direction specified. I will show this using a simple
direct proof.

1. P(x)  Q(x)
Universal Instantiation (used
twice; since for all values of x, P(x)
is true, we can pick some arbitrary
x from the universe of discourse for
which P(x) is true. Same is true for
Q(x)). Note that these are free
variables derived from universally
quantified expressions.

2. x(P(x)  Q(x))
Universal Generalization
(since an arbitrary x was used to instantiate both P(x) and Q(x), we can now make the general
statement that for all x, P(x) or Q(x) will be true. As I said before, all free variables are derived from
universal quantifiers, and not existential ones.)
Informal Explanation:
The expression above can then be translated as “If everyone is only a painter or
everyone is only a cook then everyone is either a painter or a cook.” To see how this
works, we can utilize an informal proof by cases. Let’s say that everyone in the world is
a painter, but not everyone in the world is a cook. This would readily imply that
everyone in the world is a painter or a cook, specifically, a painter. Similarly, if everyone
in the world is a cook, but not everyone in the world is a painter also implies that
everyone in the world is a painter or a cook, specifically, a cook. Or, in the case that
everyone in the world is a painter and everyone in the world is a cook, this also implies
that everyone in the world is a painter or a cook (but in this case, each and every person is
both of those things).
Counter Example:
This implication cannot hold in the opposite direction. Here is the problem that
arises. Saying simply that everyone in the world is a painter or a cook can create a
situation in which some people are painters, some are cooks, and others are both. With
this mixed environment, it is impossible to conclude that everyone in the world is a cook,
or everyone in the world is a painter, or that everyone in the world is both of those things.
So it is not the case that each x is only a painter, or each x is only a cook, or each x is
both.
PART C
x(P(x)  Q(x))  xP(x)  xQ(x)
This implication is only valid in the direction specified. To prove this, I will use an
indirect proof that will show that the negation of the conclusion does indeed imply the
negation of the hypothesis.



1. (xP(x)  xQ(x))
2. xP(x)  xQ(x)
3. xP(x)  xQ(x)
A.P. for indirect proof.
DeMorgan’s Law
Quantifier Negation (basically,
4. P(x)  Q(x)
since there does not exist some x
for which P(x) is true, not P(x) is
true for all values x. Same is true
for Q(x) and not Q(x))
Universal Instantiation (used
twice; since for all values of x, not
P(x) is true, we can pick some
arbitrary x from the universe of
discourse for which not P(x) is
true. Same is true for not Q(x)).

5. x(P(x)  Q(x))
Universal Generalization
(since we used an arbitrary x in
step 4, we can now say that for all
values of x from the universe of
discourse, not P(x) or not Q(x) will
be true.) We do not have to worry
about the free variables since all
were originally derived from
universally quantified expressions,
never an existential expression.)


6. x((P(x)  (Q(x)))
7. (x(P(x)  (Q(x)))
DeMorgan’s Law
Quantifier Negation (basically,
since for all x in the universe of
discourse, it is not the case that
P(x) and Q(x) are true, we can say
that it is not the case that for some
specific (existential) x, P(x) and
Q(x) will be true.)
8. (xP(x)  xQ(x))  (x(P(x)  (Q(x)) )
9. x(P(x)  Q(x))  xP(x)  xQ(x)
Direct Proof, steps 1-7.
By Indirect Proof.
Q.E.D.
Informal Explanantion: This implication holds in the direction specified because it is
only logical that if you find some person x that is a painter and a cook, you will be able to
again look through the universe of discourse and find a person who is a painter, and a
person who is a cook. In this case, this will be the same person (as specified in the
hypothesis).
Counterexample: On the other hand, if you find a person who is a painter, and then find
a person who is a cook, it is not necessarily one and the same person. In this case, it
would not necessarily imply that you would be able to find a person who is both a painter
and a cook.
PART D
x(P(x)  Q(x))  xP(x)  xQ(x)
The expression above is an equivalence (bi-conditional). I will show this using two
indirect proofs, first to prove that the negation of the conclusion implies the negation of
the hypothesis, and that the negation of the hypothesis implies the negation of the
conclusion.
Proving this direction: 
1. (xP(x)  xQ(x))
A.P. for indirect proof


2. xP(x)  xQ(x)
3. xP(x)  xQ(x)

4. xP(x)

5. xQ(x)

6. P(x)

7. Q(x)
DeMorgan’s Law
Quantifier Negation (basically,
since there does not exist some x
for which P(x) is true, not P(x) is
true for all values x. The same
reasoning applies to not Q(x))
3. Simplification (since we
know
that xP (x) and xQ(x) are
true, xP(x) alone must be true.)
3. Simplification (since we
know
that xP(x) and xQ(x) are
true, xQ(x) alone must be true.)
4. Universal Instantiation
(since for all x in the universe of
discourse, P(x) is true, it must be
true when we choose an arbitrary x
value.)
5. Universal Instantiation

8. P(x)  Q(x)
(since for all x in the universe of
discourse, Q(x) is true, it must be
true when we choose an arbitrary x
value.)
4,5 Conjunction since we know
that both expressions are true, the
conjunction of the two expressions
must hold true by definition of
conjunction.

7. x(P(x)  Q(x))
Universal Generalization

8. x(P(x)  Q(x))
DeMorgan’s Law

9. (x(P(x)  (Q(x)))
Quantifier Negation (basically,
(since an arbitrary x was used to instantiate both P(x) and Q(x), we can now make the general
statement that for all x, P(x) and Q(x) will be true. As I said before, all free variables are derived from
universal quantifiers, and not existential ones.)
since for all x in the universe of
discourse, it is not the case that
P(x) or Q(x) are true, we can say
that it is not the case that for some
specific (existential) x, P(x) or Q(x)
will be true.)
10. (xP(x)  xQ(x))  (x(P(x)  (Q(x)))
11. x(P(x)  Q(x))  xP(x)  xQ(x)
Direct Proof, steps 1-9.
By Indirect Proof.
Q.E.D.
x(P(x)  Q(x))  xP(x)  xQ(x)
Proving this direction: 
1. (x(P(x)  (Q(x)))
A.P. for indirect proof

2. x(P(x)  Q(x))
Quantifier Negation (since it is
not the case that for some specific
(existential) x, P(x) or Q(x) is true,
then we can assume that for all x, it
is the case that not P(x) or Q(x) is
true.)


3. x(P(x)  Q(x))
4. P(x)  Q(x)
DeMorgan’s Law
5. Universal Instantiation

5. P(x)

6. xP(x)
(since for all x in the universe of
discourse, P(x) and Q(x) is
true, it must also be true when we
choose an arbitrary x value.)
1. Simplification (since we
know
that P(x) and Q(x) are true,
P(x) alone must be true.)

8. xQ(x)
Universal Generalization
(since an arbitrary x was used to instantiate P(x), we can now make the general statement that
for all x, P(x) will be true. As I said before, all free variables are derived from universal quantifiers, and
not existential ones.)

7. Q(x)
1. Simplification (since we
know
that P(x) and Q(x) are true,
Q(x) alone must be true.)
Universal Generalization
(since an arbitrary x was used to instantiate Q(x), we can now make the general statement that
for all x, Q(x) will be true. As I said before, all free variables are derived from universal quantifiers, and
not existential ones.)

9. xP(x)  xQ(x)
6,8 Conjunction since we know
that both expressions are true, the
conjunction of the two expressions
must hold true by definition of
conjunction.

10. xP(x)  xQ(x)
Quantifier Negation (basically,
since for all x in the universe of
discourse, it is not the case that
P(x) is true and it is not the case
that Q(x) is true,, we can say that it
is not the case that for some
specific (existential) x, P(x) will be
true and it is not the case that for a
specific x Q(x) will be true.)

11. (xP(x)  xQ(x))
12. (x(P(x)  (Q(x)))  (xP(x)  xQ(x))
13. xP(x)  xQ(x)  x(P(x)  Q(x))
DeMorgan’s Law
Direct Proof, steps 1-11.
By Indirect Proof.
Q.E.D.
Informal Explanation: This is an equivalence. If some specific x is a painter or a cook,
then we can assume that by looking through the universe of discourse, we can find a
person who is a painter, or a person who is a cook (this same person). Similarly, if we
find some person who is a painter or some person (the same or not as the first) who is a
cook, we would have found some person who is a painter or a cook.
Conclusion and Extra Credit:
I have shown for the four given pairs of propositions in predicate logic, whether
they are equivalent, whether the first implies the second, or whether the second implies
the first using the methods requested. From these proofs, I am confident that the
directions of implication that are specified can be adopted as rules in future proofs.
Namely, if a universal quantifier contains a conjunction of two predicates, that quantifier
can be “distributed” to the individual predicates and the two results joined by the
conjunction. Also, if two, separate, universally quantified predicates (of the same
variable) are joined by a conjunction, the universal quantifier can be “factored out” and
made to apply to simply the conjunction of the two predicates, i.e. propositional pair a.
Similarly, as pair d shows us, if an existential quantifier contains a disjunction, the same
procedures as outlined above can be taken, with the difference being that a disjunction is
used. If, however, we are dealing with universal quantifiers and disjunctions, only two
universally quantified predicates (of the same variable) joined by that disjunction can
have their formed changed by “factoring out” the universal quantifier and leaving the
disjunction inside of it. The reverse operation is invalid. Similarly, only if an existential
quantifier contains a conjunction can the existential quantifier be “distributed” across the
predicates to yield a conjunction of two existentially quantified predicates of the same
variable. The operation cannot be reversed. These past to cases were pairs b and c,
respectfully.