Predicate Logic

The following are some notes taken during a lecture.

Limitations of Propositional Logic

If we have statements of the form:
- “All Purdue CS students are brilliant.”
- “Alice is a Purdue CS student.”
Does it follow that “Alice is brilliant?”
This is not easy to represent in propositional logic.
We need a formalism (logic) that reasons about objects, their properties, and their relations.

Basics of Predicate Logic

Predicate logic adds the following new features:
- Variables: x, y, z (variables can take values, e.g., Alice)
- Predicates: P(x), M(x) (predicates correspond to properties of variables, e.g., is_brilliant(x))
- Quantifiers (to be covered in a few slides)
Propositional functions are generalizations of propositions.
- They contain variables and a predicate, e.g., P(x)
- Variables can be replaced by elements from their domain (e.g., is_brilliant(Alice)).

Propositional Functions

Propositional functions become propositions (and have truth values) when their variables are each replaced by a value from the domain (or bound by a quantifier). For example, is_brilliant(Alice) may be T or F.
The statement P(x) is said to be the value of the propositional function P at x.

Examples:

- 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 set of integers.

- 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(3) → ¬P(-1) 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)

Quantifiers

We need quantifiers to express the meaning of English words including all and some:
- “All Purdue CS Students are brilliant.”
- “Some Purdue CS Student will win the Turing award.”
The two most important quantifiers are:
- Universal Quantifier, “For all,” symbol: ∀
- Existential Quantifier, “There exists,” symbol: ∃
We write these 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.

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)).

Translating from English into Predicate 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?
    - Hint: Consider a student Bob, who is not a student. What is S(Bob)? What is S(Bob) ∧ J(Bob)?
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 2: 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?
    - Hint: Consider the case when no student has taken a course in Java. Also that there is a student not in the class, Bob, who has not taken Java either. What is S(Bob)→ J(Bob))?
Example 3: Introduce the propositional functions Purdue_CS_Student(x) denoting “x is a Purdue CS Student” and is_brilliant(x) denoting “x is brilliant” Specify the domain as all people.
- The two statements, “All Purdue CS students are brilliant”, and “Alice is a Purdue CS student”, can be written as
  - ∀x (Purdue_CS_Student(x)→ is_brilliant(x))
  - Purdue_CS_Student (Alice)
- The conclusion is:
  - Is_brilliant(Alice)
- We will show how to automate this conclusion process later.

Equivalences in Predicate Logic

Statements involving predicates and quantifiers are logically equivalent if and only if they have the same truth value
- 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)

Quantiafiers 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.

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 Java.”
- Symbolically ¬∀x J(x) and ∃x ¬J(x) are equivalent

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”
- Symbolically ¬∃ x J(x) and ∀ x ¬J(x) are equivalent

DeMorgan’s Laws for Quantifiers

The rules for negating quantifiers are:
The reasoning in the table shows that:
These are important. You will use these.

Translating from English to Predicate 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)))

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)

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)

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))

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))

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))

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))

System Specification in Predicate Logic

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 in state x.
Now we have:

Lewis Carrol in Predicate Logic

The first two are called premises and the third is called the conclusion.
1. “All lions are fierce.”
2. “Some lions do not drink coffee.”
3. “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.

Predicate Logic: More Definitions

An assertion involving predicates and quantifiers 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 domains
- some propositional functions that can be substituted for the predicates in the assertion.
Otherwise it is unsatisfiable.

The scope of a quantifier is the part of an assertion in which variables are bound by the quantifier.

Logic Programming

Prolog (from Programming in Logic) is a programming language developed in the 1970s by researchers in artificial intelligence (AI).
Prolog programs include Prolog facts and Prolog rules.
As an example of a set of Prolog facts consider the following:

   instructor(chan, math273).
   instructor(patel, ee222).
   instructor(grossman, cs301).
   enrolled(kevin, math273).
   enrolled(juana, ee222).
   enrolled(juana, cs301).
   enrolled(kiko, math273).
   enrolled(kiko, cs301).

Here the predicates instructor(p,c) and enrolled(s,c) represent that professor p is the instructor of course c and that student s is enrolled in course c.
In Prolog, names beginning with an uppercase letter are variables.
If we have a predicate teaches(p,s) representing “professor p teaches student s,” we can write the rule:
- teaches(P,S) :- instructor(P,C), enrolled(S,C).
This Prolog rule can be viewed as equivalent to the following statement in logic (using our conventions for logical statements).
- ∀p ∀c ∀s(I(p,c) ∧ E(s,c)) → T(p,s))
Prolog programs are loaded into a Prolog interpreter. The interpreter receives queries and returns answers using the Prolog program.
For example, using our program, the following query may be given:
- ?enrolled(kevin,math273).
Prolog produces the response:
- yes
Note that the ? is the prompt given by the Prolog interpreter indicating that it is ready to receive a query.
The query:
- ?enrolled(X,math273). produces the response:
  - X = kevin; (user type ; to get next result, keeps print all answers)
  - X = kiko;
  - no
The query:
- ?teaches(X,juana). produces the response:
  - X = patel;
  - X = grossman;
  - no
The query:
- ?teaches(chan,X). produces the response:
  - X = kevin;
  - X = kiko;
  - no
There is much more to Prolog and to the entire field of logic programming.

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 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) where P(x, y) is (x + y = 0)
Nested Loops
- To see if ∀x∀yP (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 ∀x ∀yP(x,y) is false and both the outer and inner loop terminate.
  - ∀x ∀y P(x,y) is true if the outer loop ends after stepping through each x.

foreach (x in X) 
 foreach (y in Y)
  if P(x,y) == F then return false
return true

- To see if ∀x ∃yP(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.

foreach (x in X) 
 foreach (y in Y)
  if P(x,y) == T then return true
return true

Order of Quantifiers

Examples:
- Let P(x,y) be the statement “x + y = y + x.” Assume that U is the real numbers. Then ∀x ∀yP(x,y) and∀y ∀xP(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 ∃yQ(x,y) is true, but ∃y∀xQ(x,y) is false.
Example: Let U be the real numbers, and define P(x,y) : x ∙ y = 0
- What is the truth value of the following:
  - ∀x∀yP(x,y) === ∀x∀y (x . y = 0)
    - Answer: False
  - ∀x∃yP(x,y)
    - Answer: True
  - ∃x∀y P(x,y)
    - Answer: True
  - ∃x ∃ y P(x,y)
    - Answer: True
Example: Let U be the real numbers,
- Define P(x,y) : x / y = 1
- What is the truth value of the following:
  - ∀x∀yP(x,y)
    - Answer: False
  - ∀x∃yP(x,y) not every X can be divided into 1, ex x = 0
    - Answer: False
  - ∃x∀y P(x,y) same as above
    - Answer: False
  - ∃x ∃ y P(x,y)
    - Answer: True

Quantification of two tables

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 2: Translate the statement
- ∃x∀y ∀z ((F(x, y)∧ F(x,z) ∧ (y ≠z))→¬F(y,z))
- Solution: There is a student 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 Predicate Logic

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 f ” and Q(f,a) be “f is a flight on 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.

Calculus Revisited – Definition of a Limit (ε-δ Definition)

For a given function f(x), as x approaches a specific value a, we say that the limit of f(x) as x approaches a is L, denoted as:

lim_{x → a} f(x) = L

If, for every real number ε > 0, there exists a real number δ > 0 such that if the distance between x and a (i.e., |x – a|) is smaller than δ, then the distance between f(x) and L (i.e., |f(x) – L|) is smaller than ε. Mathematically, this is expressed as:

∀ ε > 0, ∃ δ > 0 such that 0 < |x – a| < δ → |f(x) – L| < ε

Here’s a breakdown of what this means:

- ε represents a small positive number (a small “margin of error”).
- δ represents another positive number.

The definition asserts that, no matter how small you choose ε to be, you can find a δ such that if x is within δ units of a, then f(x) is within ε units of L.

In more intuitive terms, it means that as you get closer and closer to a, the values of f(x) get closer and closer to L.

This definition is a fundamental concept in calculus used to rigorously define the limit of a function. It’s the basis for many limit theorems and plays a crucial role in understanding continuity, derivatives, and integrals in calculus.

Translating English to Predicate Logic

Choose the obvious predicates and express in predicate logic.
- Example 1: “Brothers are siblings.”
  - Solution: ∀x ∀y (B(x,y) → S(x,y)). // all brothers, all siblings where a brother implies sibling of 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) // all people, there exist y where loved by (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 ) ∨ ¬ Q(f,a)) by De Morgan’s for ∧.
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”

Negating Quantified Expressions: DeMorgan’s Laws

Consider again ∃ x J(x)
- “There exists a Purdue CS student who has climbed Mt. Everest.” Where J(x) is “x has climbed Mt. Everest” and the domain is all Purdue CS students.”
Negating the original statement gives “It is not the case that there is a Purdue CS student who has climbed Mt. Everest.”
Stated otherwise, “For each Purdue CS student, said student has not climbed Mt. Everest”, which is the same as ∀ x ¬J(x). We informally note from this that ¬∃ x J(x) and ∀ x ¬J(x) are equivalent

Let the domain of x be {x0, x1, x2, x3….}
We know that
- ∃ x J(x) = J(x0) ∨ J(x1) ∨ J(x2) ∨ J(x3) ∨….
Therefore,
- ¬∃ x J(x) = ¬(J(x0) ∨ J(x1) ∨ J(x2) ∨ J(x3) ∨….)
Each of the terms on the right hand side, J(x0), J(x1), J(x2), .. Are propositions. So, we can apply DeMorgan’s law for propositions to the right hand side and get:
- ¬(J(x0) ∨ J(x1) ∨ J(x2) ∨ J(x3) ∨….) = (¬ J(x0) ∧ ¬ J(x1) ∧ ¬ J(x2) ∧ ¬ J(x3)… )
Which is the same as: ∀ x ¬J(x)

Consider again the domain of x to be {x0, x1, x2, x3….}
We know that
- ∀ x J(x) = J(x0) ∧ J(x1) ∧ J(x2) ∧ J(x3) ∧.…
Therefore,
- ¬ ∀ x J(x) = ¬(J(x0) ∧ J(x1) ∧ J(x2) ∧ J(x3) ∧….)
Again, each of the terms on the right hand side, J(x0), J(x1), J(x2), .. are propositions. So, we can apply DeMorgan’s law for propositions to the right hand side and get:
- ¬(J(x0) ∧ J(x1) ∧ J(x2) ∧ J(x3) ∧….) = (¬ J(x0) ∨ ¬ J(x1) ∨ ¬ J(x2) ∨¬ J(x3)… )
Which is the same as: ∃ x ¬J(x)

DeMorgan’s Laws for Quantifiers

The rules for negating quantifiers are:
The reasoning in the table shows that:
These are important. You will use these.

Let P(x) be the proposition “x will be a millionaire”, and the domain of x be all Purdue CS students.
Question: Express the sentence “Every Purdue CS student will be a millionaire.” using quantifiers.
- Answer: ∀ x P(x)
Question: Negate the logic expression above and apply DeMorgan’s law.
- Answer:
Question: Convert this negation back into english.
- Answer: There exists some Purdue CS student who is not a millionaire.

Let P(x) be the proposition “x will win the Turing award”, and the domain of x be all Purdue CS students.
Question: Express the sentence “Some Purdue CS student will win the Turing award.” using quantifiers.
- Answer: ∃ x P(x)
Question: Negate the logic expression above and apply DeMorgan’s law.
- Answer:
Question: Convert this negation back into english.
- Answer: There does not exist a Purdue CS student who will win the Turing award. Alternately, no Purdue CS student will win the Turing award.

solidfish

Predicate Logic

Limitations of Propositional Logic

Basics of Predicate Logic

Propositional Functions

Compound Expressions

Calculus Revisited – Definition of a Limit (ε-δ Definition)

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