CS 205 Section 8, Spring 2003
John Clark
Project 1: Propositional Statements and Determination of Implication
There is no owner’s manual for the inner working of the mind. Logic is a complex transform man
comes with in which raw data, previous analysis, etc. are all inputs resulting in a conclusion. The study of
logic is the attempt to decode that transform. It is an attempt to transcend the fallibility of man in the
process of reason and record the true rules in which logic is based. A minute part of this quest is the
education of man on the rules as they have been discerned to date. This project is an application of those
rules over four specific propositional statements to determine if for the each statement the left hand side
(LHS) and right hand side (RHS) are equivalent, or if the LHS implies the RHS or if the RHS implies the
LHS.
The first propositional statement posed is x(P(x)  Q(x)) ?? xP(x)  xQ(x). This will be shown
that these statements are equivalent in meaning. The first proof will take the forward direction of the
implication, the LHS implying the RHS.
Proof:
1.
x(P(x)  Q(x))
RHS of original proposition
2.

(P(a)  Q(a))
Universal Instantiation of 1
3.

P(a)
Simp of 2
4.

Q(a)
Simp of 2
5.

xP(x)
Universal Generalization of 3
6.

xQ(x)
Universal Generalization of 4
7.

xP(x)  xQ(x)
Conj of 5,6, also equal to LHS
The simplest way to understand why this statement holds is to truly understand the  symbol as it is
implied by the ‘for all x’, e.g. x. The first part of the statement is saying that for every value x, statement
P and statement Q hold true. For a particular value of x, P and Q are both true and for another value of x, P
and Q are once again both true. This statement repeats to include all values of x in the universe of
discourse. Notice that P and Q are independent of each other now that we know they are true for every
value of x-we know this by the same conjunction (and) being used between P and Q of the same value x
and P and Q of different values of x. Having established their independence from each other in the first
part of the statement we can now say that for every value of x P(x) is true and for every value of x Q(x) is
true. This is the second part of the statement that is seen above and why this implication holds true.
The reverse direction of the equivalency is the RHS implying the LHS and this also holds true.
Proof:
1.
xP(x)  xQ(x)
LHS of original proposition
2.

xP(x)
Simp of 1
3.

xQ(x)
Simp of 1
4.

P(a)
Universal Instantiation of 2
5.

Q(a)
Universal Instantiation of 3
6.

(P(a)  Q(a))
Conj of 4,5
7.
 x(P(x)  Q(x))
Universal Generalization of 6, also equal to
RHS
Continuing the argument above discussing the  symbol, if one says that for all x something is true, the
statement can be restated as x1 is true and x2 is true and…….xn is true for all values in the universe of
discourse. If one says all of one thing is true and separately say all of another thing is true for the same
independent variable, the meaning is insensitive to saying that for a particular value of the independent
variable, one thing is true and another thing is true. To say that for all values of the independent variable
over the universe of discourse is also making the same statement and is implied meaning shown above.
The second propositional statement posed is x(P(x)  Q(x)) ?? xP(x)  xQ(x). It will be
shown that the RHS implies the LHS however the inverse is not true. The proof for the direction that holds
is shown below.
Proof:
1.
xP(x)  x Q(x)
RHS of original proposition
2.

P(a)  Q(b)
Universal Instantiation x2
3.

P(a)  Q(a)
(Clarification that these are arbitrary
numbers over the same universe of discourse)
4.

x(P(x)  Q(x))
Universal Generalization of 3, also equal to
LHS
This proof starts with knowing that either P(x) is true for all x or Q(x) is true for all x, or both are true for
all x. We can say for any arbitrary value of x, either P(x) is true (if P(x) holds for all x) or Q(x) is true (if
Q(x) holds for all x) or both are true. This can be said for any value of x since it was arbitrary so it can now
be generalized for all values of x and this is what we see in step 4 which is equivalent to the LHS of the
posed proposition.
What does not hold is the implication from the LHS to the RHS. A simple way to understand this
is to think of the universe of discourse being all positive integers. Set P(x) equal to true for all odd values
of x and set Q(x) equal to true for all even values of x. The LHS of the posed proposition would hold true
even though P(x) and Q(x) are each flip-flopping between true and false as x is incremented because they
are in opposite states in comparison to each other. If this same behavior of P(x) and Q(x) is carried to the
RHS of the posed proposition the RHS evaluates as false while the LHS is true. If P(x) and Q(x) are flip
flopping state then neither holds true for all values and universal quantification of each alone proves false.
The third propositional statement posed is x(P(x)  Q(x)) ?? xP(x)  xQ(x). The LHS implies
the RHS however the RHS does not imply the LHS. The proof for the implication that holds is shown
below.
Proof:
1.
x(P(x)  Q(x))
LHS of original proposition
2.

P(c)  Q(c)
Existential Instantiation of 1
3.

P(c)
Simp of 2
4.

Q(c)
Simp of 2
5.

xP(x)
Existential Generalization of 3
6.

xQ(x)
Existential Generalization of 4
7.

xP(x)  xQ(x)
Conj of 5,6, also equal to RHS
It is best to explain this in more common terms to fully understand the implication. The starting point is
saying that there is some value of x that the statements P(x) and Q(x) are both true for at the same time.
Now that it is known that there is a particular value of x that makes each P(x) and Q(x) true we can say that
for all values of x, there is at least one that makes P(x) true (xP(x)) and at least one that makes Q(x) true
(xQ(x)). Turning that past sentence into logical symbols the and turns into  thereby making that
sentence the same as step 7 of the above proof.
A real world example of why the RHS does not imply the LHS can be posed with the universe of
discourse consisting of one complete deck of playing cards. A single x will be a card from the deck.
Assume P(x) means x is a heart and Q(x) means x is a spade. The RHS can then be interpreted as saying
that over the whole deck of cards at least one is a heart and at least one is a spade. The LHS would be
interpreted as saying there is at least one card that is a heart and a spade at the same time. This is obviously
a contradiction and demonstrates why the implication from RHS to LHS does not hold.
The fourth propositional statement posed is x(P(x)  Q(x)) ?? xP(x)  xQ(x). The LHS and
RHS side of this statement are equivalent. To demonstrate that the LHS side implies the RHS a proof was
created (below).
Proof:
1.
x(P(x)  Q(x))
LHS of original proposition
2.

P(a)  Q(a)
Existential Instantiation of 1
3.

xP(x)  xQ(x)
Existential Generalization of 2, also equal to
RHS
or, seeing the above as applying all the correct rules but lacking more descriptive detail,
1.
x(P(x)  Q(x))
2.

LHS of original proposition
(P(x1)  Q(x1))  (P(x2)  Q(x2))  … (P(xn)  Q(xn))
defn. of
existential quantification from 1
3.

P(x1)  Q(x1)  P(x2)  Q(x2)  …  P(xn)  Q(xn) Assc. of 2
4.

(P(x1)  P(x2)  … P(xn))  (Q(x1)  Q(x2) ...  Q(xn))
Assc., Comm. of
3
5.

xP(x)  xQ(x)
defn. of existential quantification x2, also
equal to RHS
This implication holds true based on the joining piece for P and Q of the LHS being an ‘or’ as it is also on
the RHS. The nature of the existential quantifier is that it only looks for one instance to be true to make the
statement it is applied to true. This being said, the LHS is only looking for one value of x that makes either
P(x) true or Q(x) true over the entire universe of discourse of x. If only one half of the compound
statement has to be true for only one value of x for the whole LHS to be true that is the same as saying for
all of the possible values of x, there is one P(x) that equates to true or there is one Q(x) that evaluates to
true to make the entire statement true. Converting this last sentence to logical symbols, the RHS of the
posed propositional statement is found. It was already mentioned that the RHS implies the LHS, the proof
behind this is below.
Proof:
1.
xP(x)  xQ(x)
2.

RHS of original proposition
(P(x1)  P(x2)  … P(xn))  (Q(x1)  Q(x2) ...  Q(xn))
defn. of
existential quantification x2
since universe of discourse is the same for both x,
3.  P(x1)  Q(x1)  P(x2)  Q(x2)  …  P(xn)  Q(xn) Assoc., Comm of 2
4.

(P(x1)  Q(x1))  (P(x2)  Q(x2))  … (P(xn)  Q(xn))
5.

x(P(x)  Q(x))
Assoc. of 3
defn. of existential quantification, also equal
to LHS
A more informal discussion can be had as to how the RHS implies the LHS. The RHS side is symbolically
saying that over all the possible values of x, there is possibly a value of x that makes P(x) true or there is
possibly a value of x that makes Q(x) true but at least one value of one statement has to true for one value
of x over the entire universe of discourse. This statement is indifferent as to whether they both evaluate to
true for some arbitrary value each or both evaluate to true for the same value of x. It is understood that the
universe of discourse for both is the same. The definition of the inclusive or is that one or the other or both
of something must be true for a basic ‘or’ statement to be true. A few sentences back it was stated that over
a universe of discourse applied to x, P(x) is true or Q(x) is true or both are true for this statement for some
value of x. Applying the ‘or’ definition, it is not hard to understand that the implication holds ground and
the RHS implies the LHS.
The conclusions drawn in the previous analysis can be stated as
x(P(x)  Q(x))  xP(x)  xQ(x),
xP(x)  xQ(x)  x(P(x)  Q(x)),
x(P(x)  Q(x))  xP(x)  xQ(x),
and
x(P(x)  Q(x))  xP(x)  xQ(x).
Reflecting on these conclusions, additional rules that can be used as a tool in working through proofs can
be hypothesized. These will first be posed then discussed.
Rule 1: Distributive Property of Universal Quantifiers
If a compound statement, consisting of atomic propositions joined only by the  operator,
is universally quantified, then the universal quantifier can be applied to each atomic
proposition individually, and vice-versa if all propositions share the same universe of
discourse.
Rule 2: Absorptive Property of Universal Quantifiers
If a group of propositions are each universally quantified over the same universe of
discourse and joined by an  operator, then the universal quantifier can be separated out
and applied over the entire statement of the two propositions.
Rule 3: Distributive Property of Existential Quantifiers
If a compound statement, consisting of atomic propositions joined only by the  operator,
is existentially quantified, then the existential quantifier can be applied to each atomic
proposition individually, and vice-versa if all propositions share the same universe of
discourse.
Rule 4: Separative Property of Existential Quantifiers
If a compound statement, consisting of atomic propositions joined only by an  operator
is existentially quantified, then the existential quantifier can be applied to each atomic
proposition individually
Rule 1 and 3 behave like equivalence rule and rule 2 and 4 behave like inference rules. Rule 2
demonstrates that if a more constrained proposition yields (equivalent to a subset) true then the more
general application of the rule holds true, however not vice-versa as reason would state, over the entire
universe of discourse. Rule 4 is based on the same concept as rule 2, if the more constrained case can hold
true over the universe of discourse then the relaxed criteria will also pose true in the same universe of
discourse. Rules 1 and 3 are easily understood if brought into the realm of set theory, the equivalence
becomes obvious.