Predicate Logic, Universal and Existential Quantifiers


Discrete Mathematics I

7th lecture, November 15, 2019

Martin J. Dürst


© 2005-19 Martin J. Dürst Aoyama Gakuin University

Today's Schedule



There will be a minitest (ca. 30minutes) on November 29. Please start to prepare early!


Leftovers of Last Lecture

Summary of Last Lecture


Homework Due November 7, Problem 1

Prove/check the following laws using truth tables (important: Always add a sentence stating the conclusion):

  1. Reductio ad absurdum: A→¬A = ¬A
    A ¬A A→¬A
    F T T
    T F F

    Because the columns for ¬A and A→¬A are the same, this law is correct.

  2. Contraposition: AB = ¬B→¬A
    A B AB ¬B ¬A ¬B→¬A (AB) ↔(¬B→¬A)
    F F T T T T T
    F T T F T T T
    T F F T F F T
    T T T F F T T

    Because the column for (AB) ↔(¬B→¬A) is all Ts, this law is correct.

  3. The associative law for conjunction: (AB) ∧ C = A ∧  (BC)
    A B C AB (AB) ∧ C BC A ∧ (BC)
    F F F F F F F
    F F T F F F F
    F T F F F F F
    F T T F F T F
    T F F F F F F
    T F T F F F F
    T T F T F F F
    T T T T T T T

    Because the columns for (AB) ∧ C  and A ∧  (BC) are the same, this law is correct.

  4. One of De Morgan's laws: ¬(AB) = ¬A ∨ ¬B
    A B AB ¬(AB) ¬A ¬B
    ¬A ∨ ¬B
    F F F T T T T
    F T F T T F T
    T F F T F T T
    T T T F F F F

    Because the columns for ¬(AB) and ¬A ∨ ¬B are the same, this law is correct.


Homework Due November 7, Problem 2

Prove transitivity of implication (((AB) ∧ (BC)) ⇒ (AC)) by formula manipulation.

For each simplification step, indicate which law you used.

Hint: Show that ((AB) ∧ (BC)) → (AC) is a tautology by simplifying it to T.

((AB) ∧ (BC)) → (AC) [removing implication]
= ¬((AB) ∧ (BC))∨ (AC) [removing implication, 3 times]
= ¬((¬AB) ∧ (¬BC)) ∨ (¬AC) [DeMorgan]
= (¬(¬AB) ∨ ¬(¬BC)) ∨ (¬AC) [DeMorgan, twice]
= ((¬¬A∧¬B) ∨ (¬¬B∧¬C)) ∨ (¬AC) [double negation, twice]
= ((A∧¬B) ∨ (B∧¬C)) ∨ (¬AC) [associative/commutative laws]
= (A∧¬B) ∨ ¬A ∨ (B∧¬C) ∨ C [distributive law, twice]
= (A∨¬A)∧(¬B∨¬A) ∨ (BC) ∧ (¬CC) [law of excluded middle, twice]
= T ∧ (¬B∨¬A) ∨ (BC) ∧ T [property of true, twice]
= (¬B∨¬A) ∨ (BC) [associative/commutative laws]
= ¬AB∨¬BC [law of excluded middle]
= ¬A ∨ T ∨ C [property of true, twice]
= T ∴


Homework Due November 14, Problems 1/2

1. Create a set with four elements. If you use the same elements as other students, a deduction of points will be applied.

Example: {cat, cow, crow, camel}

2. Create the powerset of the set you created in problem 1.

Example: {{}, {cat}, {cow}, {crow}, {camel}, {cat, cow}, {cat, crow}, {cat, camel}, {cow, crow}, {cow, camel}, {crow, camel}, {cat, cow, crow}, {cat, cow, camel}, {cat, crow, camel}, {cow, crow, camel}, {cat, cow, crow, camel}}


Homework Due November 14, Problems 3/4

3. For sets A of size zero to six, create a table of the sizes of the powersets (|P(A)|).

|A| |P(A)|
0 1
1 2
2 4
3 8
4 16
5 32
6 64

4. Express the relationship between the size of a set A and the size of its powerset P(A) as a formula.

|P(A)| = 2|A|


Homework Due November 14, Problem 5

5. Explain the reason behind the formula in problem 4.

The formula is correct for |A|=0: A={}, P(A)={{}}, |P(A)|=20=1

If the formula is correct for |A|=k (i.e. |P(A)|=2k),
we can show that it is correct for B with |B|=k+1

Example: A={cat, cow}, |A|=2, P(A) = {{}, {cat}, {cow}, {cat, cow}}, |P(A)|=4
B = A∪{carp}={cat, cow, carp}, |B|=3, P(B) = P(A)∪{a∪{carp}|aP(A)}
=P(A)∪{{carp}, {cat, carp}, {cow, carp}, {cat, cow, carp}}

|P(B)| = 2·|P(A)| = 8

General case:

B = {c|cAc=ddA},
|B| = k+1 = |A|+1,
|P(B)|=2·|P(A)| = 2·2k=2k+1

(Let B be the set consisting of the elements of A and one additional element d which is not contained in A.
The size of B is one greater than the size of A.
Then the powerset of B is the union of two distinct sets: the powerset of A, and the set of sets from the powerset of A with d added.
The size of the powerset of B is therefore double the size of the powerset of A.)


Homework Due November 14, Problem 6

6. Create a table that shows, for sets A of size zero to five, and for each n (size of sets in P(A)), the number of such sets.

|A| n |{B|BA∧|B|=n}| |A| n |{B|BA∧|B|=n}|
0 0 1 4 0 1
1 0 1 4 1 4
1 1 1 4 2 6
2 0 1 4 3 4
2 1 2 4 4 1
2 2 1 5 0 1
3 0 1 5 1 5
3 1 3 5 2 10
3 2 3 5 3 10
3 3 1 5 4 5
5 5 1


Types of Symbolic Logic


Limitations of Propositions

With propositions, related statements have to be made separately

2 is even. 5 is even.
Today it is sunny. Tomorrow it is sunny. The day after tomorrow, it is sunny.

We can express "If today is sunny, then tomorrow will also be sunny." or "If 2 is even, then 3 is not even".

But we cannot express "If it's sunny on a given day, it's also sunny on the next day." or "If x is even, then x+2 is also even.".

⇒ This problem can be solved using predicates


Examples of Predicates


Predicate Overview


How to Write Predicates

There are two ways to write predicates:

  1. Functional notation:
  2. Operator notation:


Formulas Containing Predicates

Using predicates, we can express new things:

Similar to propositions, predicates can be true or false.

But predicates can also be unknown/undefined, for example if they contain variables.

Also, even if a predicate is undefined (e.g. even( x)),
a formula containing this predicate can have a defined value (true or false)
(e.g. even(y) → even(y+2), or odd(z) → even(z+24))


First Order Predicate Logic


Universal Quantifier

Example: ∀n∈ℕ: even(n) → even(n+2)


General form: ∀x: P (x)

∀ is the A of "for All", inverted.

Readings in Japanese:


Universal Quantifier for the Empty Set

x∈{}: P(x) = T

We have to check P(x) for all elements x in the set.
If we find just one x where P(x) is false, the overall statement is false.
But we cannot find any x where P(x) is false.
Therefore, ∀x∈{}: P(x) is always true

Application example:

All students in this room from Hungary are over 50 (years old).
sS: native(s, Hungary) → age(s) > 50

See: Vacuous truth,


Knowledge about Field of Application


Existential Quantifier

Example: ∃n∈ℕ: odd (n)


General form: ∃y: P (y)

∃ is the mirrored form of the E in "there Exists".

Readings in Japanese:


Structure of Quantifier Expressions

Example: ∀m, n∈ℕ: m > nm2n2


More Quantifier Examples

n∈ℕ: n + n + n = 3n

n∈ℕ: n2 = n3

n∈ℝ: n2 < 50n < n3

m, n∈ℕ: 7m + 2n = 2n + 7m


Applied Quantifier Examples

S: Set of students

F: Set of foods

like(p, f): Person p likes food f

All students like all foods:

Some students like all foods:

There is a food that all students like:

There is no food that all students like:

Each student dislikes a food:


Peano Axioms in Predicate Logic

Peano Axioms (Guiseppe Peano, 1858-1932)

  1. 1∈ℕ
  2. a∈ℕ: s(a)∈ℕ
  3. ¬∃x∈ℕ: s(x) = 1
  4. a, b∈ℕ: abs(a) ≠ s(b)
  5. P(1) ∧ (∀k∈ℕ: (P(k)→P(s(k)))) ⇒ ∀a∈ℕ: P(a)


The Use of Variables with Quantifiers

Bound variable:
Variable quantified by a quantifier
Example: the x in: ∀x: (P(x)∧Q(y))
Free variable:
Variable not quantified by a quantifier
Example: the y in: ∀x: (P(x)∧Q(y))
Closed formula:
A formula without free variables.
The part of a formula where a bound variable (or a quantifier) is active.
All occurrences of a bound variable within its scope can be exchanged by another variable.
Example: ∀sS: (age(s)≤30 ∧ college(s)=CSE) ⇔ ∀uS: (age(u)≤30 ∧ college(u)=CSE)
Using a bound variable outside its scope is an error.
Example: (∀x: P(x))∧Q(x)


Manipulation of Bound Variables

(∀sS: age(s)≤30) ∧ (∀tS: college(t)=CSE) = ∀uS: (age(u)≤30∧college(u)=CSE)

is the same as

(∀sS: age(s)≤30) ∧ (∀sS: college(s)=CSE) = ∀sS: (age(s)≤30∧college(s)=CSE)

There are three different variables s in the last statement.



Important Points for Quantifiers


Laws for Quantifiers

  1. ¬∀x: P(x) = ∃x: ¬P(x)
  2. ¬∃x: P(x) = ∀x: ¬P(x)
  3. (X≠{}∧∀xX: P(x)) → (∃x: P(x))
  4. (∀x: P(x)) ∧ Q(y) = ∀x: P(x)∧Q(y)
  5. (∃x: P(x)) ∧ Q(y) = ∃x: P(x)∧Q(y)
  6. (∀x: P(x)) ∨ Q(y) = ∀x: P(x)∨Q(y)
  7. (∃x: P(x)) ∨ Q(y) = ∃x: P(x)∨Q(y)
  8. (∀x: P(x)) ∧ (∀x: R(x)) = ∀x: P(x)∧R(x)
  9. (∀x: P(x)) ∨ (∀x: R(x)) ⇒ ∀x: P(x)∨R(x)
  10. (∃x: P(x)) ∨ (∃x: R(x)) = ∃x: P(x)∨R(x)
  11. (∃x: P(x)) ∧ (∃x: R(x)) ⇐ ∃x: P(x)∧R(x)
  12. (∃y: ∀x: P(x, y)) ⇒ (∀x: ∃y: P(x, y))
  13. P(x) is a tautology ⇔∀x: P(x) is a tautology


Combination of Quantifiers

(∃y: ∀x: P(x, y)) ⇒ (∀x: ∃y: P(x, y))

(∀x: ∃y: P(x, y)) ⇏ (∃y: ∀x: P(x, y))

The number of prime numbers is infinite.

(This means that whatever big number x we choose, there will always be a bigger prime number y.)

x: ∃y: (y > x ∧ prime(y))

Reversing the order of the quantifiers changes the meaning:

y: ∀x: (y > x ∧ prime(y))

(There is a prime number y that is bigger than any (natural number) x. This statement is obviously false.)


Proof that the Number of Prime Numbers is Infinite




This Week's Homework 1

Deadline: November 21, 2017 (Thursday), 19:00.

Format: Handout, easily readable handwriting

Where to submit: Box in front of room O-529 (building O, 5th floor)

Problems: See handout



predicate logic
恒真 (式)、トートロジー
恒偽 (式)
symbolic logic
multi-valued logic
fuzzy logic
first-order predicate logic
temporal logic
binary logic
higher-order logic
universal quantifier
全称限量子 (全称記号)
existential quantifier
存在限量子 (存在記号)
College of Science and Engineering
native of ...
bound variable
free variable
local variable
closed formula
prime number