Logics (cont.)
Nguyen An Khuong et.
al.
Chapter 2
Logics (cont.)
Contents
Predicate Logic
Discrete Structures for Computer Science (CO1007) on Ngày
5 tháng 1 năm 2023
Proof Methods
Some problems for
discussion
Nguyen An Khuong et. al.
Faculty of Computer Science and Engineering
University of Technology, VNU-HCM
2.1
Contents
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
1 Predicate Logic
Proof Methods
Some problems for
discussion
2 Proof Methods
3 Some problems for discussion
2.2
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3:
Some problems for
discussion
2.3
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3:
Some problems for
discussion
2.3
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3: not a proposition.
Some problems for
discussion
2.3
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3: not a proposition.
• All square numbers are not prime numbers. 100 is a square
number. Therefore 100 is not a prime number:
Some problems for
discussion
2.3
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3: not a proposition.
• All square numbers are not prime numbers. 100 is a square
number. Therefore 100 is not a prime number:
Some problems for
discussion
2.3
Limitations of Propositional Logic
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• x > 3: not a proposition.
• All square numbers are not prime numbers. 100 is a square
number. Therefore 100 is not a prime number: not able to
infer in propositional logic.
Some problems for
discussion
2.3
Predicates
Definition
Logics (cont.)
Nguyen An Khuong et.
al.
• A predicate (vị từ) is a statement containing one or more
variables.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Contexts: properties, relations, characteristics, features,...
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Contexts: properties, relations, characteristics, features,...
Notations:
• x > 3 → P (x)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Contexts: properties, relations, characteristics, features,...
Notations:
• x > 3 → P (x)
• 5 > 3 → P (5)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Contexts: properties, relations, characteristics, features,...
Notations:
• x > 3 → P (x)
• 5 > 3 → P (5)
• 2 > 3 → P (2)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Predicates
Definition
• A predicate (vị từ) is a statement containing one or more
variables.
• If values are assigned to all the variables in a predicate, the
resulting statement is a proposition (mệnh đề ).
Examples:
• x > 3 (predicate)
• 5 > 3 (proposition)
• 2 > 3 (proposition)
Etymology: predicate (n.)
from Latin praedicatum “that which is said of the
subject.”
Contexts: properties, relations, characteristics, features,...
Notations:
• x > 3 → P (x)
• 5 > 3 → P (5)
• 2 > 3 → P (2)
• A predicate with n variables P (x1 , x2 , ..., xn )
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.4
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Quantifiers:
• ∀: Universal – Với mọi
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Quantifiers:
• ∀: Universal – Với mọi
• ∀xP (x) = P (x) is T for all x
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Quantifiers:
• ∀: Universal – Với mọi
• ∀xP (x) = P (x) is T for all x
• ∃: Existential – Tồn tại
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Quantifiers:
• ∀: Universal – Với mọi
• ∀xP (x) = P (x) is T for all x
• ∃: Existential – Tồn tại
• ∃xP (x) = There exists an element x such that P (x) is T
2.5
Truth value and Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• x > 3 is true or false?
• 5>3
• For every number x, x > 3 holds
• There is a number x such that x > 3
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Quantifiers:
• ∀: Universal – Với mọi
• ∀xP (x) = P (x) is T for all x
• ∃: Existential – Tồn tại
• ∃xP (x) = There exists an element x such that P (x) is T
• We need a domain of discourse for variable: Universe
2.5
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
• P (3) = 3 < 2 is false
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
• P (3) = 3 < 2 is false
• ⇒ ∀xP (x) is false
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
• P (3) = 3 < 2 is false
• ⇒ ∀xP (x) is false
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• 3 is a counterexample (phản ví dụ) of ∀xP (x)
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
• P (3) = 3 < 2 is false
• ⇒ ∀xP (x) is false
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• 3 is a counterexample (phản ví dụ) of ∀xP (x)
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Let P (x) be the statement “x < 2”. What is the truth value of the
quantification ∀xP (x), where the domain consists of all real
number?
• P (3) = 3 < 2 is false
• ⇒ ∀xP (x) is false
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• 3 is a counterexample (phản ví dụ) of ∀xP (x)
Example
What is the truth value of the quantification ∃xP (x), where the
domain consists of all real number?
2.6
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
Some problems for
discussion
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
• M (x) = “x comes from Central Vietnam”
Some problems for
discussion
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
• M (x) = “x comes from Central Vietnam”
• Domain for x is the students in the class
Some problems for
discussion
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
• M (x) = “x comes from Central Vietnam”
• Domain for x is the students in the class
Some problems for
discussion
• ∃xM (x)
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
• M (x) = “x comes from Central Vietnam”
• Domain for x is the students in the class
Some problems for
discussion
• ∃xM (x)
Solution 2
2.7
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Express the statement “Some student in this class comes from
Central Vietnam.”
Contents
Predicate Logic
Proof Methods
Solution 1
• M (x) = “x comes from Central Vietnam”
• Domain for x is the students in the class
Some problems for
discussion
• ∃xM (x)
Solution 2
• Domain for x is all people
• ...
2.7
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
• Let C(x) denote “x is a CSE student”
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
• Let C(x) denote “x is a CSE student”
• Let S(x) denote “x studies Discrete Math 1”
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
• Let C(x) denote “x is a CSE student”
• Let S(x) denote “x studies Discrete Math 1”
• ∀x : C(x) → S(x)
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
• Let C(x) denote “x is a CSE student”
• Let S(x) denote “x studies Discrete Math 1”
• ∀x : C(x) → S(x)
• ∃x : ¬(C(x) → S(x)) ≡ ∃x : C(x) ∧ ¬S(x)
2.8
Logics (cont.)
Negation of Quantifiers
Nguyen An Khuong et.
al.
Statement
Negation
Equivalent form
∀xP (x)
¬(∀xP (x))
∃x¬P (x)
∃xP (x)
¬(∃xP (x))
∀x¬P (x)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
• All CSE students study Discrete Math 1
• Let C(x) denote “x is a CSE student”
• Let S(x) denote “x studies Discrete Math 1”
• ∀x : C(x) → S(x)
• ∃x : ¬(C(x) → S(x)) ≡ ∃x : C(x) ∧ ¬S(x)
• There is a CSE student who does not study Discrete Math 1.
2.8
Another Example
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Translate these:
• All lions are fierce.
• Some lions do not drink coffee.
• Some fierce creatures do not drink coffee.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.9
Another Example
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Translate these:
• All lions are fierce.
• Some lions do not drink coffee.
• Some fierce creatures do not drink coffee.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
Let P (x), Q(x) and R(x) be the statements “x is a lion”, “x is
fierce” and “x drinks coffee”, respectively.
• ∀x(P (x) → Q(x)).
2.9
Another Example
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Translate these:
• All lions are fierce.
• Some lions do not drink coffee.
• Some fierce creatures do not drink coffee.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
Let P (x), Q(x) and R(x) be the statements “x is a lion”, “x is
fierce” and “x drinks coffee”, respectively.
• ∀x(P (x) → Q(x)).
• ∃x(P (x) ∧ ¬R(x)).
2.9
Another Example
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Translate these:
• All lions are fierce.
• Some lions do not drink coffee.
• Some fierce creatures do not drink coffee.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
Let P (x), Q(x) and R(x) be the statements “x is a lion”, “x is
fierce” and “x drinks coffee”, respectively.
• ∀x(P (x) → Q(x)).
• ∃x(P (x) ∧ ¬R(x)).
• ∃x(Q(x) ∧ ¬R(x)).
2.9
The Order of Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• The order of quantifiers is important, unless all the quantifiers
are universal quantifiers or all are existential quantifiers
Contents
Predicate Logic
Proof Methods
Example
Some problems for
discussion
∀x ∀y (x + y = y + x)
T for all x, y ∈ R
2.10
The Order of Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• The order of quantifiers is important, unless all the quantifiers
are universal quantifiers or all are existential quantifiers
• Read from left to right, apply from inner to outer
Example
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∀x ∀y (x + y = y + x)
T for all x, y ∈ R
2.10
The Order of Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• The order of quantifiers is important, unless all the quantifiers
are universal quantifiers or all are existential quantifiers
• Read from left to right, apply from inner to outer
Example
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∀x ∀y (x + y = y + x)
T for all x, y ∈ R
2.10
The Order of Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• The order of quantifiers is important, unless all the quantifiers
are universal quantifiers or all are existential quantifiers
• Read from left to right, apply from inner to outer
Example
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∀x ∀y (x + y = y + x)
T for all x, y ∈ R
Example
∀x ∃y (x + y = 0) is T,
2.10
The Order of Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
• The order of quantifiers is important, unless all the quantifiers
are universal quantifiers or all are existential quantifiers
• Read from left to right, apply from inner to outer
Example
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∀x ∀y (x + y = y + x)
T for all x, y ∈ R
Example
∀x ∃y (x + y = 0) is T, while
∃y ∀x (x + y = 0) is F
2.10
Quantifiers plus ∧ and ∨
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Show ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x). (That is, show that
no matter what the domain is, these 2 propositions always have
the same truth value).
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.11
Quantifiers plus ∧ and ∨
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Show ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x). (That is, show that
no matter what the domain is, these 2 propositions always have
the same truth value).
Contents
Predicate Logic
Proof Methods
Terminology: We say that ∀ distributes over ∧.
Some problems for
discussion
2.11
Quantifiers plus ∧ and ∨
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Show ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x). (That is, show that
no matter what the domain is, these 2 propositions always have
the same truth value).
Contents
Predicate Logic
Proof Methods
Terminology: We say that ∀ distributes over ∧.
Some problems for
discussion
Chứng minh.
First assume that ∀x(P (x) ∧ Q(x)) is true. So for all x, P (x) is
true and Q(x) is true. Therefore ∀xP (x) is true, and ∀xQ(x) is
true. Therefore ∀xP (x) ∧ ∀xQ(x) is true.
2.11
Quantifiers plus ∧ and ∨
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Show ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x). (That is, show that
no matter what the domain is, these 2 propositions always have
the same truth value).
Contents
Predicate Logic
Proof Methods
Terminology: We say that ∀ distributes over ∧.
Some problems for
discussion
Chứng minh.
First assume that ∀x(P (x) ∧ Q(x)) is true. So for all x, P (x) is
true and Q(x) is true. Therefore ∀xP (x) is true, and ∀xQ(x) is
true. Therefore ∀xP (x) ∧ ∀xQ(x) is true.
Now assume ∀xP (x) ∧ ∀xQ(x) is true. So ∀xP (x) is true and
∀xQ(x) is true. So for all x, P (x) is true and for all x, Q(x) is
true. Therefore, for all x, P (x) ∧ Q(x) is true. So
∀x(P (x) ∧ Q(x)) is true.
2.11
Quantifiers plus ∧ and ∨
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Show ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x). (That is, show that
no matter what the domain is, these 2 propositions always have
the same truth value).
Contents
Predicate Logic
Proof Methods
Terminology: We say that ∀ distributes over ∧.
Some problems for
discussion
Chứng minh.
First assume that ∀x(P (x) ∧ Q(x)) is true. So for all x, P (x) is
true and Q(x) is true. Therefore ∀xP (x) is true, and ∀xQ(x) is
true. Therefore ∀xP (x) ∧ ∀xQ(x) is true.
Now assume ∀xP (x) ∧ ∀xQ(x) is true. So ∀xP (x) is true and
∀xQ(x) is true. So for all x, P (x) is true and for all x, Q(x) is
true. Therefore, for all x, P (x) ∧ Q(x) is true. So
∀x(P (x) ∧ Q(x)) is true.
Therefore ∀x(P (x) ∧ Q(x)) ≡ ∀xP (x) ∧ ∀xQ(x).
2.11
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
3
With ∨, the situation is reversed. ∃ distributes over ∨, but ∀
does not.
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
3
With ∨, the situation is reversed. ∃ distributes over ∨, but ∀
does not.
a. ∃x(S(x) ∨ R(x)) ≡ ∃xS(x) ∨ ∃xR(x).
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
3
With ∨, the situation is reversed. ∃ distributes over ∨, but ∀
does not.
a. ∃x(S(x) ∨ R(x)) ≡ ∃xS(x) ∨ ∃xR(x).
b. ∀x(S(x) ∨ R(x)) ←− ∀xS(x) ∨ ∀xR(x) is true.
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
3
With ∨, the situation is reversed. ∃ distributes over ∨, but ∀
does not.
a. ∃x(S(x) ∨ R(x)) ≡ ∃xS(x) ∨ ∃xR(x).
b. ∀x(S(x) ∨ R(x)) ←− ∀xS(x) ∨ ∀xR(x) is true.
b. ∀x(S(x) ∨ R(x)) ̸≡ ∀xS(x) ∨ ∀xR(x)
2.12
Exercises∗
Logics (cont.)
Nguyen An Khuong et.
al.
1
2
The existential quantifier ∃ does not distribute over ∧. That
is,
∃x(P (x) ∧ Q(x)) ̸≡ ∃xP (x) ∧ ∃xQ(x).
The following is true though:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x(P (x) ∧ Q(x)) −→ ∃xP (x) ∧ ∃xQ(x).
3
With ∨, the situation is reversed. ∃ distributes over ∨, but ∀
does not.
a. ∃x(S(x) ∨ R(x)) ≡ ∃xS(x) ∨ ∃xR(x).
b. ∀x(S(x) ∨ R(x)) ←− ∀xS(x) ∨ ∀xR(x) is true.
b. ∀x(S(x) ∨ R(x)) ̸≡ ∀xS(x) ∨ ∀xR(x)
4
∃ does not distribute over −→. I.e.,
∃x(P (x) −→ Q(x)) ̸≡ ∃xP (x) −→ ∃xQ(x).
2.12
Translating Nested Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
∀x (C(x) ∨ ∃y (C(y) ∧ F (x, y)) )
Provided that:
• C(x): x has a computer,
Proof Methods
Some problems for
discussion
• F (x, y): x and y are friends,
• x, y ∈ all students in your school.
2.13
Translating Nested Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
∀x (C(x) ∨ ∃y (C(y) ∧ F (x, y)) )
Provided that:
• C(x): x has a computer,
Proof Methods
Some problems for
discussion
• F (x, y): x and y are friends,
• x, y ∈ all students in your school.
Answer
For every student x in your school, x has a computer or there is a
student y such that y has a computer and x and y are friends.
2.13
Translating Nested Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
∃x∀y∀z (((F (x, y) ∧ F (x, z) ∧ (y ̸= z)) → ¬F (y, z)))
Provided that:
• F (x, y): x, y are friends
Proof Methods
Some problems for
discussion
• x, y, z ∈ all students in your school.
2.14
Translating Nested Quantifiers
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
∃x∀y∀z (((F (x, y) ∧ F (x, z) ∧ (y ̸= z)) → ¬F (y, z)))
Provided that:
• F (x, y): x, y are friends
Proof Methods
Some problems for
discussion
• x, y, z ∈ all students in your school.
Answer
There is a student x, so that for every student y, every student z
not the same as y, if x and y are friends, and x and z are friends,
then y and z are not friends.
2.14
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
1
“There is a student in the class has visited Hanoi”.
2
“Every students in the class have visited Nha Trang or Vung
Tau”.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.15
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
1
“There is a student in the class has visited Hanoi”.
2
“Every students in the class have visited Nha Trang or Vung
Tau”.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Answer
Assume:
C(x) : x has visited Hanoi
D(x) : x has visited Nha Trang
E(x) : x has visited Vung Tau
We have:
2.15
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
1
“There is a student in the class has visited Hanoi”.
2
“Every students in the class have visited Nha Trang or Vung
Tau”.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Answer
Assume:
C(x) : x has visited Hanoi
D(x) : x has visited Nha Trang
E(x) : x has visited Vung Tau
We have:
1
∃xC(x)
2.15
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
1
“There is a student in the class has visited Hanoi”.
2
“Every students in the class have visited Nha Trang or Vung
Tau”.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Answer
Assume:
C(x) : x has visited Hanoi
D(x) : x has visited Nha Trang
E(x) : x has visited Vung Tau
We have:
1
∃xC(x)
2
∀x(D(x) ∨ E(x))
2.15
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Example
Proof Methods
Every people has one and only one best friend.
Some problems for
discussion
2.16
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Example
Proof Methods
Every people has one and only one best friend.
Some problems for
discussion
Solution
Assume:
• B(x, y) : y is the best friend of x
We have:
2.16
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Example
Proof Methods
Every people has one and only one best friend.
Some problems for
discussion
Solution
Assume:
• B(x, y) : y is the best friend of x
We have: ∀x∃y∀z(B(x, y) ∧ ((y ̸= z) → ¬B(x, z)))
2.16
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
If a person is a woman and a parent, then this person is mother of
some one.
Proof Methods
Some problems for
discussion
2.17
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
If a person is a woman and a parent, then this person is mother of
some one.
Proof Methods
Some problems for
discussion
Solution
We define:
• C(x) : x is woman
• D(x) : x is a parent
• E(x, y): x is mother of y
We have:
2.17
Translating into Logical Expressions
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
Predicate Logic
If a person is a woman and a parent, then this person is mother of
some one.
Proof Methods
Some problems for
discussion
Solution
We define:
• C(x) : x is woman
• D(x) : x is a parent
• E(x, y): x is mother of y
We have: ∀x((C(x) ∧ D(x)) → ∃yE(x, y))
2.17
Inference
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• If I have a girlfriend, I will take her to go shopping.
• Whenever I and my girlfriend go shopping and that day is a
special day, I will surely buy her some expensive gift.
• If I buy my girlfriend expensive gifts, I will eat noodles for a
week.
• Today is March 8.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• March 8 is such a special day.
• Therefore, if I have a girlfriend,...
2.18
Inference
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• If I have a girlfriend, I will take her to go shopping.
• Whenever I and my girlfriend go shopping and that day is a
special day, I will surely buy her some expensive gift.
• If I buy my girlfriend expensive gifts, I will eat noodles for a
week.
• Today is March 8.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• March 8 is such a special day.
• Therefore, if I have a girlfriend,...
• I will eat noodles for a week.
2.18
Propositional Rules of Inferences
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
p
p→q
∴q
Name
Contents
Predicate Logic
Proof Methods
Modus ponens
Some problems for
discussion
2.19
Propositional Rules of Inferences
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
p→q
∴q
Modus ponens
¬q
p→q
∴ ¬p
Modus tollens
Predicate Logic
Proof Methods
Some problems for
discussion
2.19
Propositional Rules of Inferences
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
p→q
∴q
Modus ponens
¬q
p→q
∴ ¬p
Modus tollens
p→q
q→r
∴p→r
Hypothetical syllogism
(Tam đoạn luận giả định)
Predicate Logic
Proof Methods
Some problems for
discussion
2.19
Propositional Rules of Inferences
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
p→q
∴q
Modus ponens
¬q
p→q
∴ ¬p
Modus tollens
p→q
q→r
∴p→r
Hypothetical syllogism
(Tam đoạn luận giả định)
p∨q
¬p
∴q
Disjunctive syllogism
(Tam đoạn luận tuyển)
Predicate Logic
Proof Methods
Some problems for
discussion
2.19
Logics (cont.)
Propositional Rules of Inferences
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.20
Logics (cont.)
Propositional Rules of Inferences
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
∴p∨q
Addition
(Quy tắc cộng )
Predicate Logic
Proof Methods
Some problems for
discussion
2.20
Logics (cont.)
Propositional Rules of Inferences
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
∴p∨q
p∧q
∴p
Addition
(Quy tắc cộng )
Predicate Logic
Proof Methods
Some problems for
discussion
Simplification
(Rút gọn)
2.20
Logics (cont.)
Propositional Rules of Inferences
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
∴p∨q
Addition
(Quy tắc cộng )
p∧q
∴p
Simplification
(Rút gọn)
p
q
∴p∧q
Conjunction
(Kết hợp)
Predicate Logic
Proof Methods
Some problems for
discussion
2.20
Logics (cont.)
Propositional Rules of Inferences
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
p
∴p∨q
Addition
(Quy tắc cộng )
p∧q
∴p
Simplification
(Rút gọn)
p
q
∴p∧q
Conjunction
(Kết hợp)
p∨q
¬p ∨ r
∴q∨r
Resolution
(Phân giải)
Predicate Logic
Proof Methods
Some problems for
discussion
2.20
Logics (cont.)
Nguyen An Khuong et.
al.
Example
If it rains today, then we will not have a barbecue today. If we do
not have a barbecue today, then we will have a barbecue
tomorrow. Therefore, if it rains today, then we will have a
barbecue tomorrow.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.21
Logics (cont.)
Nguyen An Khuong et.
al.
Example
If it rains today, then we will not have a barbecue today. If we do
not have a barbecue today, then we will have a barbecue
tomorrow. Therefore, if it rains today, then we will have a
barbecue tomorrow.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
• p: It is raining today
• q: We will not have a barbecue today
• r: We will have barbecue tomorrow
2.21
Logics (cont.)
Nguyen An Khuong et.
al.
Example
If it rains today, then we will not have a barbecue today. If we do
not have a barbecue today, then we will have a barbecue
tomorrow. Therefore, if it rains today, then we will have a
barbecue tomorrow.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
• p: It is raining today
• q: We will not have a barbecue today
• r: We will have barbecue tomorrow
p→q
q→r
∴p→r
Hypothetical syllogism
2.21
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• It is not sunny this afternoon
(¬p) and it is colder than
yesterday (q)
• We will go swimming (r)
only if it is sunny
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• If we do not go swimming,
then we will take a canoe trip
(s)
• If we take a canoe trip, then
we will be home by sunset (t)
2.22
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• It is not sunny this afternoon
(¬p) and it is colder than
yesterday (q)
• We will go swimming (r)
only if it is sunny
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• If we do not go swimming,
then we will take a canoe trip
(s)
• If we take a canoe trip, then
we will be home by sunset (t)
• We will be home by sunset
(t)
2.22
Logics (cont.)
Nguyen An Khuong et.
al.
Example
1. ¬p ∧ q
Hypothesis
2. ¬p
Simplification using (1)
3. r → p
Hypothesis
4. ¬r
Modus tollens using (2) and (3)
5. ¬r → s
Hypothesis
• If we take a canoe trip, then
we will be home by sunset (t)
6. s
Modus ponens using (4) and (5)
• We will be home by sunset
(t)
7. s → t
Hypothesis
8. t
Modus ponens using (6) and (7)
• It is not sunny this afternoon
(¬p) and it is colder than
yesterday (q)
• We will go swimming (r)
only if it is sunny
• If we do not go swimming,
then we will take a canoe trip
(s)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.22
Fallacies
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Definition
Fallacies (ngụy biện) resemble rules of inference but are based on
contingencies rather than tautologies.
Predicate Logic
Proof Methods
Some problems for
discussion
2.23
Fallacies
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Definition
Fallacies (ngụy biện) resemble rules of inference but are based on
contingencies rather than tautologies.
Predicate Logic
Proof Methods
Some problems for
discussion
Example
If you do correctly every questions in mid-term exam, you will get
10 grade. You got 10 grade.
Therefore, you did correctly every questions in mid-term exam.
Is [(p → q) ∧ q] → p a tautology?
2.23
Rules of Inference for Quantified Statements
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
Predicate Logic
∀xP (x)
∴ P (c)
Universal instantiation
(Cụ thể hóa phổ quát)
Proof Methods
Some problems for
discussion
2.24
Rules of Inference for Quantified Statements
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
Predicate Logic
∀xP (x)
∴ P (c)
Universal instantiation
(Cụ thể hóa phổ quát)
P (c)for an arbitrary c
∴ ∀xP (x)
Universal generalization
(Tổng quát hóa phổ quát)
Proof Methods
Some problems for
discussion
2.24
Rules of Inference for Quantified Statements
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
Predicate Logic
∀xP (x)
∴ P (c)
Universal instantiation
(Cụ thể hóa phổ quát)
P (c)for an arbitrary c
∴ ∀xP (x)
Universal generalization
(Tổng quát hóa phổ quát)
∃xP (x)
∴ P (c)for some element c
Existential instantiation
(Cụ thể hóa tồn tại)
Proof Methods
Some problems for
discussion
2.24
Rules of Inference for Quantified Statements
Logics (cont.)
Nguyen An Khuong et.
al.
Rule of Inference
Name
Contents
Predicate Logic
∀xP (x)
∴ P (c)
Universal instantiation
(Cụ thể hóa phổ quát)
P (c)for an arbitrary c
∴ ∀xP (x)
Universal generalization
(Tổng quát hóa phổ quát)
∃xP (x)
∴ P (c)for some element c
Existential instantiation
(Cụ thể hóa tồn tại)
P (c)for some element c
∴ ∃xP (x)
Existential generalization
(Tổng quát hóa tồn tại)
Proof Methods
Some problems for
discussion
2.24
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• A student in this class has not gone to class
• Everyone in this class passed the first exam
Contents
• Someone who passed the first exam has not gone to class
Some problems for
discussion
Predicate Logic
Proof Methods
2.25
Logics (cont.)
Nguyen An Khuong et.
al.
Example
• A student in this class has not gone to class
• Everyone in this class passed the first exam
Contents
• Someone who passed the first exam has not gone to class
Some problems for
discussion
Predicate Logic
Proof Methods
Hint
• C(x): x is in this class
• B(x): x has gone to class
• P (x): x passed the first exam
• Premises???
2.25
Logics (cont.)
Nguyen An Khuong et.
al.
1. ∃x(C(x) ∧ ¬B(x))
Premise
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1. ∃x(C(x) ∧ ¬B(x))
2. C(a) ∧ ¬B(a)
Premise
Existential instantiation from (1)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1. ∃x(C(x) ∧ ¬B(x))
2. C(a) ∧ ¬B(a)
3. C(a)
Premise
Existential instantiation from (1)
Simplification from (2)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
5.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
C(a) → P (a)
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Universal instantiation from (4)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
5.
6.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
C(a) → P (a)
P (a)
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Universal instantiation from (4)
Modus ponens from (3) and (5)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
5.
6.
7.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
C(a) → P (a)
P (a)
¬B(a)
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Universal instantiation from (4)
Modus ponens from (3) and (5)
Simplification from (2)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
5.
6.
7.
8.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
C(a) → P (a)
P (a)
¬B(a)
P (a) ∧ ¬B(a)
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Universal instantiation from (4)
Modus ponens from (3) and (5)
Simplification from (2)
Conjunction from (6) and (7)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Logics (cont.)
Nguyen An Khuong et.
al.
1.
2.
3.
4.
5.
6.
7.
8.
9.
∃x(C(x) ∧ ¬B(x))
C(a) ∧ ¬B(a)
C(a)
∀x(C(x) → P (x))
C(a) → P (a)
P (a)
¬B(a)
P (a) ∧ ¬B(a)
∃x(P (x) ∧ ¬B(x))
Premise
Existential instantiation from (1)
Simplification from (2)
Premise
Universal instantiation from (4)
Modus ponens from (3) and (5)
Simplification from (2)
Conjunction from (6) and (7)
Existential generalization from (8)
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.26
Introduction
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Definition
Some problems for
discussion
A proof is a sequence of logical deductions from
- axioms, and
- previously proved theorems
that concludes with a new theorem.
2.27
Terminology
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• Theorem (định lý ) = a statement that can be shown to be
true
• Axiom (tiên đề ) = a statement we assume to be true
• Hypothesis (giả thiết) = the premises of the theorem
2.28
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• Lemma (bổ đề ) = less important theorem that is helpful in
the proofs of other results
• Corollary (hệ quả ) = a theorem that can be established
directly from a proved theorem
• Conjecture (phỏng đoán) = statement being proposed to be
true, when it is proved, it becomes theorem
2.29
Proving a Theorem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Many theorem has the form ∀xP (x) → Q(x)
Goal:
• Show that P (c) → Q(c) is true with arbitrary c of the domain
Proof Methods
Some problems for
discussion
2.30
Proving a Theorem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Many theorem has the form ∀xP (x) → Q(x)
Goal:
• Show that P (c) → Q(c) is true with arbitrary c of the domain
• Apply universal generalization
Proof Methods
Some problems for
discussion
2.30
Proving a Theorem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Many theorem has the form ∀xP (x) → Q(x)
Goal:
• Show that P (c) → Q(c) is true with arbitrary c of the domain
• Apply universal generalization
Proof Methods
Some problems for
discussion
2.30
Proving a Theorem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Many theorem has the form ∀xP (x) → Q(x)
Goal:
• Show that P (c) → Q(c) is true with arbitrary c of the domain
• Apply universal generalization
Proof Methods
Some problems for
discussion
⇒ How to show that conditional statement p → q is true.
2.30
Methods of Proof
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
• Direct proofs (chứng minh trực tiếp)
• Proof by contraposition (chứng minh phản đảo)
• Proof by contradiction (chứng minh phản chứng )
Some problems for
discussion
• Mathematical induction (quy nạp toán học)
2.31
Direct Proofs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Definition
A direct proof shows that p → q is true by showing that if p is
true, then q must also be true.
Predicate Logic
Proof Methods
Some problems for
discussion
2.32
Direct Proofs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Definition
A direct proof shows that p → q is true by showing that if p is
true, then q must also be true.
Predicate Logic
Proof Methods
Some problems for
discussion
Example
Ex.: If n is an odd integer, then n2 is odd.
Pr.: Assume that n is odd. By the definition, n = 2k + 1, k ∈ Z.
n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 is an odd
number.
2.32
Proof by Contraposition
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p → q can be proved by showing (directly) that its contrapositive,
¬q → ¬p, is true.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.33
Proof by Contraposition
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p → q can be proved by showing (directly) that its contrapositive,
¬q → ¬p, is true.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
Ex.: If n is an integer and 3n + 2 is odd, then n is odd.
2.33
Proof by Contraposition
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p → q can be proved by showing (directly) that its contrapositive,
¬q → ¬p, is true.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Example
Ex.: If n is an integer and 3n + 2 is odd, then n is odd.
Pr.: Assume that “If 3n + 2 is odd, then n is odd” is false; or n is
even, so n = 2k, k ∈ Z. Substituting
3n + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) is even. Because
the negation of the conclusion of the conditional statement
implies that the hypothesis is false, Q.E.D.
2.33
Proofs by Contradiction
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p is true if if can show that ¬p → (r ∧ ¬r) is true for some
proposition r.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.34
Proofs by Contradiction
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p is true if if can show that ¬p → (r ∧ ¬r) is true for some
proposition r.
Predicate Logic
Proof Methods
Some problems for
discussion
Example
Ex.: Prove that
Contents
√
2 is irrational.
2.34
Proofs by Contradiction
Logics (cont.)
Nguyen An Khuong et.
al.
Definition
p is true if if can show that ¬p → (r ∧ ¬r) is true for some
proposition r.
Predicate Logic
Proof Methods
Some problems for
discussion
Example
Ex.: Prove that
Contents
√
2 is irrational.
√
Pr.: Let p is the proposition
“ 2 is irrational”. Suppose
¬p is true,
√
√
which means 2 is rational. If so, ∃a, b ∈ Z, 2 = a/b, a, b
have no common factors. Squared, 2 = a2 /b2 , 2b2 = a2 , so
a2 is even, and a is even, too. Because of that a = 2c, c ∈ Z.
Thus, 2b2 = 4c2 , or b2 = 2c2 , which means b2 is even and so
is b. That means 2 divides both a and b, contradict with the
assumption.
2.34
Problem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Assume that we have an infinite domino string, we want to know
whether every dominoes will fall, if we only know two things:
1
We can push the first domino to fall
2
If a domino falls, the next one will be fall
2.35
Problem
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Assume that we have an infinite domino string, we want to know
whether every dominoes will fall, if we only know two things:
1
We can push the first domino to fall
2
If a domino falls, the next one will be fall
We can! Mathematical induction.
2.35
Mathematical Induction
Logics (cont.)
Nguyen An Khuong et.
al.
Definition (Induction)
To prove that P (n) is true for all positive integers n, where P (n)
is a propositional function, we complete two steps:
• Basis Step: Verify that P (1) is true.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• Inductive Step: Show that the conditional statement
P (k) → P (k + 1) is true for all positive integers k
Logic form:
[P (1) ∧ ∀kP (k) → P (k + 1))] → ∀nP (n)
2.36
Mathematical Induction
Logics (cont.)
Nguyen An Khuong et.
al.
Definition (Induction)
To prove that P (n) is true for all positive integers n, where P (n)
is a propositional function, we complete two steps:
• Basis Step: Verify that P (1) is true.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
• Inductive Step: Show that the conditional statement
P (k) → P (k + 1) is true for all positive integers k
Logic form:
[P (1) ∧ ∀kP (k) → P (k + 1))] → ∀nP (n)
What is P (n) in domino string case?
2.36
Logics (cont.)
Example on Induction
Nguyen An Khuong et.
al.
Example
Show that if n is a positive integer, then
1 + 2 + ... + n =
n(n + 1)
.
2
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.37
Logics (cont.)
Example on Induction
Nguyen An Khuong et.
al.
Example
Show that if n is a positive integer, then
1 + 2 + ... + n =
n(n + 1)
.
2
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
Let P (n) be the proposition that sum of first n is n(n + 1)/2
• Basis Step: P (1) is true, because 1 =
1(1+1)
2
2.37
Logics (cont.)
Example on Induction
Nguyen An Khuong et.
al.
Example
Show that if n is a positive integer, then
1 + 2 + ... + n =
n(n + 1)
.
2
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
Solution
Let P (n) be the proposition that sum of first n is n(n + 1)/2
• Basis Step: P (1) is true, because 1 =
• Inductive Step:
Assume that 1 + 2 + . . . + k =
Then:
1(1+1)
2
k(k+1)
.
2
1 + 2 + . . . + k + (k + 1)
=
=
=
k(k + 1)
+ (k + 1)
2
k(k + 1) + 2(k + 1)
2
(k + 1)(k + 2)
2
shows that P (k + 1) is true under the assumption that P (k) is true.
2.37
Example on Induction
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Prove that n < 2n for all positive integers n.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.38
Example on Induction
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Prove that n < 2n for all positive integers n.
Contents
Predicate Logic
Solution
Let P (n) be the proposition that n > 2n .
• Basis Step: P (1) is true, because 1 > 21 = 2
Proof Methods
Some problems for
discussion
2.38
Example on Induction
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Prove that n < 2n for all positive integers n.
Contents
Predicate Logic
Proof Methods
Solution
Let P (n) be the proposition that n > 2n .
• Basis Step: P (1) is true, because 1 > 21 = 2
• Inductive Step:
Assume that P (k) is true for the positive k, that is, k < 2k .
Add 1 to both side of k < 2k , note that 1 ≤ 2k .
Some problems for
discussion
k + 1 < 2k + 1 ≤ 2k + 2k = 2 · 2k = 2k+1 .
shows that P (k + 1) is true, namely, that k + 1 < 2k+1 ,
based on the assumption that P (k) is true.
2.38
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
• I.e. is it true independent of the domain (here pubs, people)
and the meanings of pub and to drink?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
• I.e. is it true independent of the domain (here pubs, people)
and the meanings of pub and to drink?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
• I.e. is it true independent of the domain (here pubs, people)
and the meanings of pub and to drink?
• Predicate formula:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
• I.e. is it true independent of the domain (here pubs, people)
and the meanings of pub and to drink?
• Predicate formula:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x ∈ P, [D(x) −→ ∀y ∈ P, D(y)].
2.39
On drinking in pubs
Logics (cont.)
Nguyen An Khuong et.
al.
The drinker’s paradox
In every non-empty pub there is somebody such that if he (or she)
drinks then everybody drinks.
• Is this true?
• Or more precisely: Is this a tautology in classical predicate
logic?
• I.e. is it true independent of the domain (here pubs, people)
and the meanings of pub and to drink?
• Predicate formula:
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
∃x ∈ P, [D(x) −→ ∀y ∈ P, D(y)].
• Law of excluded middle (LEM): p ∨ ¬p is a tautology.
2.39
Some MCQs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
MCQ1
Proof Methods
Which of the following has truth value T ?
Some problems for
discussion
2.40
Some MCQs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
MCQ1
Proof Methods
Which of the following has truth value T ?
Some problems for
discussion
A. ∀x ∈ R, (x > 1 −→ x2 − 3x + 2 > 0).
2.40
Some MCQs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
MCQ1
Proof Methods
Which of the following has truth value T ?
Some problems for
discussion
A. ∀x ∈ R, (x > 1 −→ x2 − 3x + 2 > 0).
B. ∃x ∈ Q, (x2 = 2015).
2.40
Some MCQs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
MCQ1
Proof Methods
Which of the following has truth value T ?
Some problems for
discussion
A. ∀x ∈ R, (x > 1 −→ x2 − 3x + 2 > 0).
B. ∃x ∈ Q, (x2 = 2015).
C. ∃x ∈ R, (x > 2 −→ x2 − 3x + 2 < 0).
2.40
Some MCQs
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Predicate Logic
MCQ1
Proof Methods
Which of the following has truth value T ?
Some problems for
discussion
A. ∀x ∈ R, (x > 1 −→ x2 − 3x + 2 > 0).
B. ∃x ∈ Q, (x2 = 2015).
C. ∃x ∈ R, (x > 2 −→ x2 − 3x + 2 < 0).
D. ∃x ∈ R, (x2 − x = −1).
2.40
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
MCQ2
Giả sử D(x, y) là một vị từ với ý nghĩa “số nguyên y là một ước
của số nguyên x.” Phát biểu nào dưới đây tương đương diễn đạt ý
nghĩa của công thức
∀x, y(D(x, y) −→ ∃z(D(x, z) ∧ D(y, z)))?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
2.41
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
MCQ2
Giả sử D(x, y) là một vị từ với ý nghĩa “số nguyên y là một ước
của số nguyên x.” Phát biểu nào dưới đây tương đương diễn đạt ý
nghĩa của công thức
∀x, y(D(x, y) −→ ∃z(D(x, z) ∧ D(y, z)))?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
A. Mọi cặp số tự nhiên (x, y) đều có ít nhất một ước chung.
2.41
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
MCQ2
Giả sử D(x, y) là một vị từ với ý nghĩa “số nguyên y là một ước
của số nguyên x.” Phát biểu nào dưới đây tương đương diễn đạt ý
nghĩa của công thức
∀x, y(D(x, y) −→ ∃z(D(x, z) ∧ D(y, z)))?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
A. Mọi cặp số tự nhiên (x, y) đều có ít nhất một ước chung.
B. Nếu y là một ước của x và z là một ước của y thì z cũng là
ước của x.
2.41
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
MCQ2
Giả sử D(x, y) là một vị từ với ý nghĩa “số nguyên y là một ước
của số nguyên x.” Phát biểu nào dưới đây tương đương diễn đạt ý
nghĩa của công thức
∀x, y(D(x, y) −→ ∃z(D(x, z) ∧ D(y, z)))?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
A. Mọi cặp số tự nhiên (x, y) đều có ít nhất một ước chung.
B. Nếu y là một ước của x và z là một ước của y thì z cũng là
ước của x.
C. Nếu y không phải là ước của x thì chúng không có ước
chung.
2.41
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
MCQ2
Giả sử D(x, y) là một vị từ với ý nghĩa “số nguyên y là một ước
của số nguyên x.” Phát biểu nào dưới đây tương đương diễn đạt ý
nghĩa của công thức
∀x, y(D(x, y) −→ ∃z(D(x, z) ∧ D(y, z)))?
Contents
Predicate Logic
Proof Methods
Some problems for
discussion
A. Mọi cặp số tự nhiên (x, y) đều có ít nhất một ước chung.
B. Nếu y là một ước của x và z là một ước của y thì z cũng là
ước của x.
C. Nếu y không phải là ước của x thì chúng không có ước
chung.
D. Nếu x và y không có ước chung thì y không phải là một ước
của x.
2.41
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
MCQ3
Which of the following are semantically and syntactically correct
translations of “No dog bites a child of its owner”?
Predicate Logic
Proof Methods
Some problems for
discussion
2.42
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
MCQ3
Which of the following are semantically and syntactically correct
translations of “No dog bites a child of its owner”?
Predicate Logic
Proof Methods
Some problems for
discussion
A. ∀xDog(x) −→ ¬Bites(x, Child(Owner(x))).
2.42
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
MCQ3
Which of the following are semantically and syntactically correct
translations of “No dog bites a child of its owner”?
Predicate Logic
Proof Methods
Some problems for
discussion
A. ∀xDog(x) −→ ¬Bites(x, Child(Owner(x))).
B. ¬∃x, ∃yDog(x) ∧ Child(y, Owner(x)) ∧ Bites(x, y).
2.42
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
MCQ3
Which of the following are semantically and syntactically correct
translations of “No dog bites a child of its owner”?
Predicate Logic
Proof Methods
Some problems for
discussion
A. ∀xDog(x) −→ ¬Bites(x, Child(Owner(x))).
B. ¬∃x, ∃yDog(x) ∧ Child(y, Owner(x)) ∧ Bites(x, y).
C. ∀xDog(x) −→ (∀yChild(y, Owner(x)) −→ ¬Bites(x, y)).
2.42
Some MCQs (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
MCQ3
Which of the following are semantically and syntactically correct
translations of “No dog bites a child of its owner”?
Predicate Logic
Proof Methods
Some problems for
discussion
A. ∀xDog(x) −→ ¬Bites(x, Child(Owner(x))).
B. ¬∃x, ∃yDog(x) ∧ Child(y, Owner(x)) ∧ Bites(x, y).
C. ∀xDog(x) −→ (∀yChild(y, Owner(x)) −→ ¬Bites(x, y)).
D. ¬∃xDog(x) −→ (∃yChild(y, Owner(x)) ∧ Bites(x, y)).
2.42
Convert Codes to English and Predicate Formula
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
for (i=0; i<numObjects; i++)
{
Object x = Objects(i);
if isMushroom(i)
if isPoisonous(x) && isPurple(x)
return false;
}
return true;
Predicate Logic
Proof Methods
Some problems for
discussion
2.43
Convert Codes to English and Predicate Formula
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
for (i=0; i<numObjects; i++)
{
Object x = Objects(i);
if isMushroom(i)
if isPoisonous(x) && isPurple(x)
return false;
}
return true;
Predicate Logic
Proof Methods
Some problems for
discussion
• There are no mushrooms that are poisonous and purple.
2.43
Convert Codes to English and Predicate Formula
Logics (cont.)
Nguyen An Khuong et.
al.
Example
Contents
for (i=0; i<numObjects; i++)
{
Object x = Objects(i);
if isMushroom(i)
if isPoisonous(x) && isPurple(x)
return false;
}
return true;
Predicate Logic
Proof Methods
Some problems for
discussion
• There are no mushrooms that are poisonous and purple.
• ∀xM ushroom(x) −→ ¬(P oisonous(x) ∧ P urple(x)).
2.43
Convert Codes to English and Predicate Formula (cont’d)
Logics (cont.)
Nguyen An Khuong et.
al.
Contents
Example
Predicate Logic
Proof Methods
for (i=0; i<numObjects; i++)
{
Object x = Objects(i);
if isMushroom(i) && isPoisonous(x) && isPurple(x)
return true;
}
return false;
Some problems for
discussion
2.44