Proofs Homework
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chapter1p2.pptxThe Foundations: Logic and Proofs
Chapter 1, Part II: Predicate and Relational Logic
Predicates, Relations and Quantifiers
Section 1.4
Propositional Logic Not Enough
If we have:
“All men are mortal.”
“Socrates is a man.”
Does it follow that “Socrates is mortal?”
Can’t be represented in propositional logic. Need a language that talks about objects, their properties, and their relations.
Later we’ll see how to draw inferences.
Introducing Predicate Logic
Predicate logic uses the following new features:
Variables: x, y, z
Predicates/Relations: P(x), M(x), Q(x,y), R(x,y,z)
Quantifiers (to be covered in a few slides)
Constants: a, b, c, ..., 0, -1, 4, Socrates, ...
Propositional functions are a generalization of propositions.
They contain variables and a predicate (property or relation), e.g., P(x), R(x,y), S(x,y,z), ...
Variables can be replaced by elements from their domain (constants) to create propositions (that are T or F)
Propositional Functions
Propositional functions become propositions (and have truth values) when their variables are each replaced by a value from the domain (or are bound by a quantifier, as we will see later).
The statement P(x) is said to be the value of the propositional function P at x.
For example, let P(x) denote “x > 0” and the domain be the integers. Then:
P(−3) is false.
P(0) is false.
P(3) is true.
Often the domain is denoted by U. So in this example U is the integers.
Examples of Propositional Functions
Let “x + y = z” be denoted by R(x, y, z) and U (for all three variables) be the integers. Find these truth values:
R(2,-1,5)
Solution: F
R(3,4,7)
Solution: T
R(x, 3, z)
Solution: Not a Proposition
Now let “x − y = z” be denoted by Q(x, y, z), with U as the integers. Find these truth values:
Q(2,-1,3)
Solution: T
Q(3,4,7)
Solution: F
Q(x, 3, z)
Solution: Not a Proposition
Compound Expressions
Connectives from propositional logic carry over to predicate logic.
If P(x) denotes “x > 0,” find these truth values:
P(3) ∨ P(-1) Solution: T
P(3) ∧ P(-1) Solution: F
P(3) → P(-1) Solution: F
P(-1) → P(3) Solution: T
Expressions with variables are not propositions and therefore do not have truth values. For example,
P(3) ∧ P(y)
P(x) → P(y)
When used with quantifiers (to be introduced next), these expressions (propositional functions) become propositions.
Quantifiers
We need quantifiers to express the meaning of English words including all and some:
“All men are Mortal.”
“Some cats do not have fur.”
The two most important quantifiers are:
Universal Quantifier, “For all,” symbol:
Existential Quantifier, “There exists,” or "Some," symbol:
We write as in x P(x) and x P(x).
x P(x) asserts P(x) is true for every x in the domain.
x P(x) asserts P(x) is true for some x in the domain.
The quantifiers are said to bind the variable x in these expressions.
Charles Peirce (1839-1914)
Universal Quantifier
x P(x) is read as “For all x, P(x)” or “For every x, P(x)”
Examples:
If P(x) denotes “x > 0” and U is the integers, then x P(x) is false.
If P(x) denotes “x > 0” and U is the positive integers, then x P(x) is true.
If P(x) denotes “x is even” and U is the integers, then
x P(x) is false.
Existential Quantifier
x P(x) is read as “For some x, P(x)”, or as “There is an x such that P(x),” or “For at least one x, P(x).”
Examples:
If P(x) denotes “x > 0” and U is the integers, then x P(x) is true. It is also true if U is the positive integers.
If P(x) denotes “x < 0” and U is the positive integers, then x P(x) is false.
If P(x) denotes “x is even” and U is the integers, then
x P(x) is true.
Uniqueness Quantifier
!x P(x) means that P(x) is true for one and only one x in the universe of discourse.
This is commonly expressed in English in the following equivalent ways:
“There is a unique x such that P(x).”
“There is one and only one x such that P(x)”
Examples:
If P(x) denotes “x + 1 = 0” and U is the integers, then !x P(x) is true.
But if P(x) denotes “x > 0,” then !x P(x) is false.
The uniqueness quantifier is not really needed as the restriction that there is a unique x such that P(x) can be expressed as: x (P(x) ∧y (P(y) → y =x))
Thinking about Quantifiers
When the domain of discourse is finite, we can think of quantification as looping through the elements of the domain.
To evaluate x P(x) loop through all x in the domain.
If at every step P(x) is true, then x P(x) is true.
If at a step P(x) is false, then x P(x) is false and the loop terminates.
To evaluate x P(x) loop through all x in the domain.
If at some step, P(x) is true, then x P(x) is true and the loop terminates.
If the loop ends without finding an x for which P(x) is true, then x P(x) is false.
Even if the domains are infinite, we can still think of the quantifiers this fashion, but the loops will not terminate in some cases.
Properties of Quantifiers
The truth value of x P(x) and x P(x) depend on both the propositional function P(x) and on the domain U.
Examples:
If U is the positive integers and P(x) is the statement “x < 2”, then x P(x) is true, but x P(x) is false.
If U is the negative integers and P(x) is the statement “x < 2”, then both x P(x) and x P(x) are true.
If U consists of 3, 4, and 5, and P(x) is the statement “x > 2”, then both x P(x) and x P(x) are true. But if P(x) is the statement “x < 2”, then both x P(x) and x P(x) are false.
Precedence of Quantifiers
The quantifiers and have higher precedence than all the logical operators.
For example, x P(x) ∨ Q(x) means (x P(x))∨ Q(x)
x (P(x) ∨ Q(x)) means something different.
Unfortunately, often people write x P(x) ∨ Q(x) when they mean x (P(x) ∨ Q(x)).
Aristotle's Square of Opposition
x (S(x)→ P(x))
x (S(x)→ ¬P(x)) OR
¬x(S(x) ∧ P(x))
x(S(x) ∧ P(x))
x(S(x) ∧ ¬P(x))
Translating from English to Logic
Example 1: Translate the following sentence into predicate logic: “Every student in this class has taken a course in Java.”
Solution:
First decide on the domain U.
Solution 1: If U is all students in this class, define a propositional function J(x) denoting “x has taken a course in Java” and translate as x J(x).
Solution 2: But if U is all people, also define a propositional function S(x) denoting “x is a student in this class” and translate as x (S(x)→ J(x)).
x (S(x) ∧ J(x)) is not correct. What does it mean?
Translating from English to Logic
Example 2: Translate the following sentence into predicate logic: “Some student in this class has taken a course in Java.”
Solution:
First decide on the domain U.
Solution 1: If U is all students in this class, translate as
x J(x)
Solution 1: But if U is all people, then translate as x (S(x) ∧ J(x))
x (S(x)→ J(x)) is not correct. What does it mean?
Returning to the Socrates Example
Introduce the propositional functions Man(x) denoting “x is a man” and Mortal(x) denoting “x is mortal.” Specify the domain as all people.
The two premises are:
The conclusion is:
Later we will show how to prove that the conclusion follows from the premises.
Equivalences in Predicate Logic
Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value (the biconditional is a tautology)
for every predicate substituted into these statements and
for every domain of discourse used for the variables in the expressions.
The notation S ≡T indicates that S and T are logically equivalent.
Example: x ¬¬S(x) ≡ x S(x)
19
Thinking about Quantifiers as Conjunctions and Disjunctions
If the domain is finite, a universally quantified proposition is equivalent to a conjunction of propositions without quantifiers and an existentially quantified proposition is equivalent to a disjunction of propositions without quantifiers.
If U consists of the integers 1,2, and 3:
Even if the domains are infinite, you can still think of the quantifiers in this fashion, but the equivalent expressions without quantifiers will be infinitely long.
20
Negating Quantified Expressions
Consider x J(x)
“Every student in your class has taken a course in Java.”
Here J(x) is “x has taken a course in Java” and
the domain is students in your class.
Negating the original statement gives “It is not the case that every student in your class has taken Java.” This implies that “There is a student in your class who has not taken calculus.”
Symbolically ¬x J(x) and x ¬J(x) are equivalent
Negating Quantified Expressions (continued)
Now Consider x J(x)
“There is a student in this class who has taken a course in Java.”
Where J(x) is “x has taken a course in Java.”
Negating the original statement gives “It is not the case that there is a student in this class who has taken Java.” This implies that “Every student in this class has not taken Java” or "No student in the class has taken Java"
Symbolically ¬ x J(x) and x ¬J(x) are equivalent
De Morgan’s Laws for Quantifiers
The rules for negating quantifiers are:
The reasoning in the table shows that:
These are important. You will use these.
Translation from English to Logic
Examples:
“Some student in this class has visited Mexico.”
Solution: Let M(x) denote “x has visited Mexico” and S(x) denote “x is a student in this class,” and U be all people.
x (S(x) ∧ M(x))
“Every student in this class has visited Canada or Mexico.”
Solution: Add C(x) denoting “x has visited Canada.”
x (S(x)→ (M(x)∨C(x)))
Some Fun with Translating from English into Logical Expressions
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
Translate “Everything is a fleegle”
Solution: x F(x)
Translation (cont)
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
“Nothing is a snurd.”
Solution: ¬x S(x) What is this equivalent to?
Solution: x ¬ S(x)
Translation (cont)
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
“All fleegles are snurds.”
Solution: x (F(x)→ S(x))
Translation (cont)
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
“Some fleegles are thingamabobs.”
Solution: x (F(x) ∧ T(x))
Translation (cont)
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
“No snurd is a thingamabob.”
Solution: ¬x (S(x) ∧ T(x)) What is this equivalent to?
Solution: x (¬S(x) ∨ ¬T(x))
Translation (cont)
U = {fleegles, snurds, thingamabobs}
F(x): x is a fleegle
S(x): x is a snurd
T(x): x is a thingamabob
“If any fleegle is a snurd then it is also a thingamabob.”
Solution: x ((F(x) ∧ S(x))→ T(x)) or
x ((F(x) → (S(x)→ T(x))
System Specification Example
Predicate logic is used for specifying properties that systems must satisfy.
For example, translate into predicate logic:
“Every mail message larger than one megabyte will be compressed.”
“If a user is active, at least one network link will be available.”
Decide on predicates and domains (left implicit here) for the variables:
Let L(m, y) be “Mail message m is larger than y megabytes.”
Let C(m) denote “Mail message m will be compressed.”
Let A(u) represent “User u is active.”
Let S(n, x) represent “Network link n is state x.
Now we have:
Lewis Carroll Example
The first two are called premises and the third is called the conclusion.
“All lions are fierce.”
“Some lions do not drink coffee.”
“Some fierce creatures do not drink coffee.”
Here is one way to translate these statements to predicate logic. Let P(x), Q(x), and R(x) be the propositional functions “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))
Later we will see how to prove that the conclusion follows from the premises.
Charles Lutwidge Dodgson
(AKA Lewis Carroll)
(1832-1898)
Some Predicate Calculus Definitions
An assertion involving predicates and quantifiers is a tautology (is valid) if it is true
for all domains
every propositional function substituted for the predicates in the assertion.
Example:
An assertion involving predicates is satisfiable if it is true
for some domain
some propositional functions that can be substituted for the predicates in the assertion.
Otherwise it is unsatisfiable.
Example: not a tautology but satisfiable
Example: unsatisfiable
More Predicate Calculus Definitions
A variable is free if is not bound by a quantifier
x is free in: F(x), R(x,y)
x is bound in: x P(x) and x P(x)
The scope of a quantifier is the part of an assertion in which variables are bound by the quantifier.
Example: x has wide scope
Example: x has narrower scope
In – Class Exercises
§1.4: #s 12, 18a-c, 24a & b, 32a, b, d, 36
Nested Quantifiers
Section 1.5
Nested Quantifiers
Nested quantifiers are often necessary to express the meaning of sentences in English as well as important concepts in computer science and mathematics.
Example: “Every real number has an additive inverse” is
x y(x + y = 0)
where the domains of x and y are the real numbers.
We can also think of nested propositional functions:
x y(x + y = 0) can be viewed as x Q(x) where Q(x) is y P(x, y) and P(x, y) is (x + y = 0)
Thinking of Nested Quantification
Nested Loops
To see if xy P (x,y) is true, loop through the values of x :
At each step, loop through the values for y.
If for some pair of x and y, P(x,y) is false, then xy P(x, y) is false and both the outer and inner loops terminate.
x y P(x,y) is true if the outer loop ends after stepping through each x.
To see if x y P(x, y) is true, loop through the values of x:
At each step, loop through the values for y.
The inner loop ends when a pair x and y is found such that P(x, y) is true.
If no y is found such that P(x, y) is true the outer loop terminates as x yP(x, y) has been shown to be false.
x y P(x,y) is true if the outer loop ends after stepping through each x.
If the domains of the variables are infinite, then this process can not actually be carried out.
Order of Quantifiers
Examples:
Let P(x,y) be the statement “x + y = y + x.” Assume that U is the real numbers. Then x y P(x,y) and y x P(x,y) have the same truth value.
Let Q(x,y) be the statement “x + y = 0.” Assume that U is the real numbers. Then x y Q(x,y) is true, but y x Q(x,y) is false.
Questions on Order of Quantifiers
Example 1: Let U be the real numbers,
Define P(x,y) : x ∙ y = 0
What is the truth value of the following:
x y P(x,y)
Answer: False
x y P(x,y)
Answer: True
xy P(x,y)
Answer: True
x y P(x,y)
Answer: True
Questions on Order of Quantifiers
Example 2: Let U be the real numbers,
Define P(x,y) : x / y = 1
What is the truth value of the following:
x y P(x,y)
Answer: False
x y P(x,y)
Answer: True
x y P(x,y)
Answer: False
x y P(x,y)
Answer: True
Quantifications with Two Variables
| Statement | When True? | When False |
| P(x,y) is true for every pair x,y. | There is a pair x, y for which P(x,y) is false. | |
| For every x there is a y for which P(x,y) is true. | There is an x such that P(x,y) is false for every y. | |
| There is an x for which P(x,y) is true for every y. | For every x there is a y for which P(x,y) is false. | |
| There is a pair x, y for which P(x,y) is true. | P(x,y) is false for every pair x,y |
Translating Nested Quantifiers into English
Example 1: Translate the statement
x (C(x )∨ y (C(y ) ∧ F(x, y)))
where C(x) is “x has a computer,” and F(x,y) is “x and y are friends,” and the domain for both x and y consists of all students in your school.
Solution: Every student in your school has a computer or has a friend who has a computer.
Example 1: Translate the statement (tough one!)
xy z ((F(x, y)∧ F(x,z) ∧ (y ≠z))→¬F(y,z))
Solution: There is some person none of whose friends are also friends with each other.
Translating Mathematical Statements into Predicate Logic
Example : Translate “The sum of two positive integers is always positive” into a logical expression.
Solution:
Rewrite the statement to make the implied quantifiers and domains explicit:
“For every two integers, if these integers are both positive, then the sum of these integers is positive.”
Introduce the variables x and y, and specify the domain, to obtain:
“For all positive integers x and y, x + y is positive.”
The result is:
x y (((x > 0)∧ (y > 0)) → (x + y > 0))
where the domain of both variables consists of all integers
Translating English into Logical Expressions Example
Example: Use quantifiers to express the statement “There is a woman who has taken a flight on every airline in the world.”
Solution:
Let P(w,f) be “w has taken flight f ” and Q(f,a) be “f is a flight on airline a .”
The domain of w is all women, the domain of f is all flights, and the domain of a is all airlines.
Then the statement can be expressed as:
w a f (P(w,f ) ∧ Q(f,a))
Calculus in Logic
Example: Use quantifiers to express the definition of the limit of a real-valued function f(x) of a real variable x at a point a in its domain.
Solution: Recall the definition of the statement
is “For every real number ε > 0, there exists a real number δ > 0 such that |f(x) – L| < ε whenever 0 < |x –a| < δ.”
Using quantifiers:
Where the domain for the variables ε and δ consists of all positive real numbers and the domain for x consists of all real numbers.
Translation from English Examples
Choose the obvious predicates and express in predicate logic.
Example 1: “Brothers are siblings.”
Solution: x y (B(x,y) → S(x,y))
Example 2: “Siblinghood is symmetric.”
Solution: x y (S(x,y) → S(y,x))
Example 3: “Everybody loves somebody.”
Solution: x y L(x,y)
Example 4: “There is someone who is loved by everyone.”
Solution: y x L(x,y)
Example 5: “There is someone who loves someone.”
Solution: x y L(x,y)
Example 6: “Everyone loves himself”
Solution: x L(x,x)
Negating Nested Quantifiers
Example 1: Recall the logical expression developed three slides back:
w a f (P(w,f ) ∧ Q(f,a))
Part 1: Use quantifiers to express the statement that “There does not exist a woman who has taken a flight on every airline in the world.”
Solution: ¬w a f (P(w,f ) ∧ Q(f,a))
Part 2: Now use De Morgan’s Laws to move the negation as far inwards as possible.
Solution:
¬w a f (P(w,f ) ∧ Q(f,a))
w ¬ a f (P(w,f ) ∧ Q(f,a)) by De Morgan’s for
w a ¬ f (P(w,f ) ∧ Q(f,a)) by De Morgan’s for
w a f ¬ (P(w,f ) ∧ Q(f,a)) by De Morgan’s for
w a f (¬P(w,f ) v ¬Q(f,a)) by De Morgan’s for ∧
Negating Nested Quantifiers
w a f (¬ P(w,f ) ∨ ¬ Q(f,a))
Part 3: Can you translate the result back into English?
Solution:
“For every woman there is an airline such that for all flights, this woman has not taken that flight or that flight is not on this airline
Return to Calculus and Logic
Example : Recall the logical expression developed in the calculus example three slides back.
Use quantifiers and predicates to express that does not exist.
We need to say that for all real numbers L,
The result from the previous example can be negated to yield:
Now we can repeatedly apply the rules for negating quantified expressions:
(see Implication Laws of propositional logic)
The last step uses the equivalence ¬(p→q) ≡ p∧¬q
Calculus in Predicate Logic
Therefore, to say that the limit does not exist:
means that:
Remember that ε and δ range over all positive real numbers and x over all real numbers.
Translating back into English we have, for every real number L, there is a real number ε > 0, such that for every real number δ > 0, there exists a real number x such that 0 < | x – a | < δ and |f(x) – L | ≥ ε .
Some Questions about Quantifiers
Can you switch the order of quantifiers?
Is this a valid equivalence?
Solution: Yes! The left and the right side will always have the same truth value. The order in which x and y are picked does not matter.
Is this a valid equivalence?
Solution: No! The left and the right side may have different truth values for some propositional functions for P. Try “x + y = 0” for P(x,y) with U being the integers. The order in which the values of x and y are picked does matter.
Can you distribute quantifiers over logical connectives?
Is this a valid equivalence?
Solution: Yes! The left and the right side will always have the same truth value no matter what propositional functions are denoted by P(x) and Q(x).
Is this a valid equivalence?
Solution: No! The left and the right side may have different truth values. Pick “x is a fish” for P(x) and “x has scales” for Q(x) with the domain of discourse being all animals. Then the left side is false, because there are some fish that do not have scales. But the right side is true since not all animals are fish.
In – Class Exercises
§1.5: #s 4 (in idiomatic English), 10a-c – use e = Evelyn, and f = Fred, 24a & b, 32d
