MATH 705
Set Theory and Logic
Module 5
Q
UANTIFICATION RULES
Objectives
After studying this module, you should be able to:
1. explain the concept of propositional function;
2. apply the ways of quantifying propositional function;
3. classify the types of subject–predicate propositional functions ;
4. show the negation of the quantified propositional function; and
5. prove the validity or invalidity of the quantified propositional function.
Introduction
This module builds the vocabulary of logic specifically on quantification theory. It tells
us the meaning of propositional functions and the way they are quantified and negated. It
also shows how quantified propositional function can be verified valid or invalid using the
methods of proofs or the shortened truth table method.
Propositional Functions
A propositional function is a general proposition which attributes a property or a
relation to unspecified elements of a set. It is an expression p(x) which has the property that
p(a) is true or false for each a  A. That is, p(x) becomes a statement (with truth value)
whenever any element a  A is substituted for the variable x.
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Example:
Find the truth set of each proposition p(x) defined on the set Q+ = {1, 2, 3, . . .}.
a) Let p(x) be “x + 2 > 7”. Then the truth set is the set
{xx  Q+, x + 2 > 7} = { 6, 7, 8, . . . } consisting of all integers greater than 5.
b) Let p(x) be “x + 5 < 3”. Then the truth set is the set
{xx  P, x + 5 < 3} =  the empty set . In other words, p(x) is not true for any
positive integer in P.
c) Let p(x) be “X + 5 > 1”. Then the truth set is the set
{xx  P, x + 5 > 1 } = P. Thus p(x) is true for every element in P.
Propositional function can be generated using the following:
1. If P is a propositional function, then ~P is also a propositional function.
2. If P and Q are propositional functions then so are P  Q, P  Q, P  Q and P  Q.
Ways of Quantifying a Propositional Function
1. Universal quantification is the assertion “for all x, P(x)” denoted by xP(x). x is said to
be universally quantified.
Example:
Let U = R = {xx is a real number}. If P(x) denoted the proposition “x2 > 0”, then P(x) is
valid in U, i.e. xP(x) is true.
Interpretation:
a) For all x, the assertion P(x) is true.
b) If the universe of discourse is U, this assertion xP(x) is true iff the predicate P is
valid in U.
c) If c is any element of U, the implication xP(x)  P(c) is true.
d) Let U = {a1, a2, a3, . . . , an}. xP(x) is materially equivalent to P(a1)  P(a2) 
P(a3)  . . .  P(an).
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Set Theory and Logic
2. Existential quantification is the assertion is the assertion “for some x, P(x)” denoted by
x P(x). x is said to be existentially quantified.
Example:
Let P(x) be the proposition “4x – 3 < 0” then P(x) is satisfied in U, i.e. x P(x) is true.
Interpretation:
a) For some x, the assertion P(x) is true.
b) If the universe of discourse is U, this assertion x P(x) is true iff P(x) is satisfied
in U.
c) It follows that for an element c in U, the implication P(c)  x P(x) is true.
d) Let U = {a1, a2, a3, . . . , an}. xP(x) is materially equivalent to P(a1)  P(a2)  P(a3)
 . . .  P(an).
Wait! … pause for a while, answer first the following question.
SAQ1
Let U = Z = {xx is an integer}. Tell whether the following propositions are true or false.
Justify your answer.
1. x[10x2 + 7x – 12 = 0]
2. x[x + 4 < 7]
3. xy[3y  x]
4. xy[x + y = 10]
ASAQ1
1. x[10x2 + 7x – 12 = 0] is false since the roots are not integers.
2. x[x + 4 < 7] is true since there are values of x that will satisfy the equation that is
when x < 3 or (-, 3)
3. xy[3y  x] is false since 2y can be equal to x when y = 1 and x = 3.
4. xy[x + y = 10] is true since there are combinations of integers that satisfy this
equation.
Four Types of Subject-Predicate Propositions of Traditional Logic
1. Universal Affirmative: x[P(x)  Q(x)]
2. Universal Negative:
x[P(x)  ~Q(x)]
3. Particular Affirmative: x[P(x)  Q(x)]
4. Particular Negative:
Module 5: Quantification Rules
x[P(x)  ~Q(x)]
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Set Theory and Logic
Example:
Symbolize the following propositions.
1. All boys took the exam.
x[B(x)  T(x)]
2. No boys failed the exam.
x[B(x)  ~F(x)]
3. Some boys are present.
x[B(x)  P(x)]
4. Some boys are not attending their classes. x[B(x)  ~A(x)]
5. Ice cream and cake are sweet.
x{[I(x)  C(x)]  S(x)}
6. All materials are either old or newly purchased.
x{M(x)  [O(x)  N(x)]
7. Some instructors are either temporary or permanent.x{I(x)  [T(x)  P(x)]}
8. The only even prime number is 2.
x{[E(x)  P(x)]  x = 2}
Wait! … pause for a while, answer first the following question.
SAQ2
Given the following general propositions:
All students are humble.
Some students are pretty.
No students are humble.
Some students are ugly.
a) Symbolize the propositions using the given representations.
b) Instantiate using the following universe of discourse.
i)
U = {Keith}
ii)
U = {Keith, Lorenz}
ASAQ2
1. x[S(x)  D(x)]
i)
S(Keith)  D(Keith)
ii) [S(Keith)  D(Keith)]  [S(Lorenzs)  D(Lorenz)]
2. x[S(x)  P(x)]
i)
S(Keith)  P(Keith)
ii) [S(Keith)  P(Keith)]  [S(Lorenz)  P(Lorenz)]
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Set Theory and Logic
3. x[S(x)  ~D(x)]
i)
S(Keith)  ~D(Keith)
ii) [S(Keith)  ~D(Keith)]  [S(Lorenz)  ~D(LOrenz)]
4. x[S(x)  ~P(x)]
i)
S(Keith)  ~P(Keith)
ii) [S(Keith)  ~P(Keith)]  [S(Lorenz)  ~P(Lorenz)]
Quantification Negation
1. ~x P(x)  x ~P(x)
2. ~x P(x)  x ~P(x)
3. ~x [P(x)  Q(x)]  x [P(x)  ~Q(x)]
4. ~x [P(x)  Q(x)]  x[P(x)  ~Q(x)]
Example:
State the negation of the following propositions.
1. All students returned the books.
This can be represented by x[S(x)  R(x)]. Based on rule 3, the negation of this
proposition is x[S(x)  ~R(x)]. Thus, the negation of the given proposition is: Some
students did not return the books.
2. Some students played the sepak takraw game.
This can be represented by x[S(x)  P(x)]. Based on rule 4, the negation of this
proposition is x[S(x) ~W(x)]. Thus, the negation of the given proposition is: No
student played the sepak takraw game.
Proving the Invalidity of Arguments Involving Quantifiers
The proof of invalidity of arguments involving quantifiers also involves finding
conditions when the premises are true and the conclusion is false. Notice that this condition
becomes complicated when we deal with general propositions since all the elements in the
universe of discourse have to be considered.
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Set Theory and Logic
Let U = {C1, C2, C3,. . ., Cn} be the universe of discourse.
We interpret the following quantified proposition to mean
xP(x)  P(C1)  P(C2)  P(C3)  . . .  P(Cn)
xP(x)  P(C1)  P(C2)  P(C3)  . . .  P(Cn)
Example:
Show that the following argument is invalid.
[x P(x)  x Q(x)]  x [P(x)  Q(x)]
U = {a, b}
{[P(a)  P(b)]  [Q(a)  Q(b)]}  {[P(a)  P(a)]  [Q(b)  Q(b)]}
Assigning the following truth values :
P(a) P(b) Q(a) Q(b)
,
,
,
T
F
F
T
Quantification Rules
Let A(x) be a well-formed formula with a free occurrence of x such that no free
occurrence of x in A(x) is within the scope of y or y.
1. Universal Instantiation (U.I.)
∀xA ( x )
∴ A(y)
2. Existential Instantiation (E.I.)
∃xA(x)
∴ A(y)
3. Existential Generalization (E.G.)
A(x)
∴ ∃yA(y)
4. Universal Generalization (U.G.)
A(x)
∴ ∀yA(y)
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Set Theory and Logic
Example:
Construct a formal proof validity of the following arguments.
a)
1. x{A(x)  ~[B(x)  C(x)]}
2. A(s)
/ A(s)  ~B(s)
3. A(s)  ~[B(s)  C(s)] 1 U.I.
4. ~[B(s)  C(s)]
2,3 MP
5. ~B(s)  ~C(s)
4 De M.
6. ~B(s)
5 Simp.
7. A(s)  ~B(s)
2-6 C.P.
b)
1. x[T(x) [F(x)  D(x)]
2. x[T(x)  B(x)]
/x[D(x)  B(x)]
3. T(a) [F(a)  D(a)]
1 U.I.
4. T(a)  B(a)
2 E.I.
5. T(a)
4 Simp.
6. B(a)
4 Comm., Simp.
7. F(a)  D(a)
3,5 M.P.
8. D(a)
7 Comm., Simp.
9. D(a)  B(a)
8,6 Conj.
10. x[D(x)  B(x)]
9 E.G.
Wait! … pause for a while, answer first the following question.
SAQ3
Prove
1. x{[W(x)  S(x)]  E(x)}
2. x[S(x)  W(x)]
/x[S(x)  E(x)]
ASAQ3
1. x{[W(x)  S(x)]  E(x)}
2. x[S(x)  W(x)]
/x[S(x)  E(x)]
3. S(y)
Assumption
4. S(y)  W(y)
2 U.I.
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Set Theory and Logic
5. [W(y)  S(y)]  E(y)
1 U.I.
6. W(y)
4,3 M.P.
7. W(y)  S(y)
6, 3 Conj.
8. E(y)
5,7 M.P.
9. S(y)  E(y)
3-8 C.P.
10. x[S(x)  E(x)]
9 U.G.
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Activity No. 5
1. Consider the following argument form involving quantifiers:
∃x[B(x) ∧ C(x)
∀x[A(x) → C(x)
∴ ∀x[B(x) → A(x)
a. Let B(x) stand for being brave; C(x) for being compassionate; and A(x) for being
affectionate. State the corresponding propositions of the propositional functions in
the given argument.
b. Instantiate the given propositional functions using a universe with one element.
c. Find out if the resulting argument is invalid by the shortened truth table method.
d. Instantiate the given propositional functions using a universe with two elements.
e. Find out if the resulting argument is invalid by the shortened truth table.
2. Construct the formal proof of validity.
∀x{N(x) → [R(x) ∨ M(x)]}
∀x[R(x) → A(x)
∃x[N(x)∧ ~ A(x)]
∴ ∃x[N(x)∧ M(x)]
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