1. Logical Equivalence & Predicate Logic
이재원
School of Information Technique
Sungshin W. University
2008-fall

2. Propositional Equivalence
Two syntactically (i.e., textually) different compound propositions may be the semantically identical (i.e., have the same meaning). We call them equivalent. Learn:
Various equivalence rules or laws.
How to prove equivalences using symbolic derivations.

3. Tautologies and Contradictions
A tautology is a compound proposition that is true no matter what the truth values of its atomic propositions are!
Ex. p  p [What is its truth table?]
A contradiction is a compound proposition that is false no matter what! Ex. p  p [Truth table?]
Other compound props. are contingencies.

4. Logical Equivalence
Compound proposition p is logically equivalent to compound proposition q, written pq, IFF the compound proposition pq is a tautology.
Compound propositions p and q are logically equivalent to each other IFF p and q contain the same truth values as each other in all rows of their truth tables.

5. Proving Equivalence via Truth Tables
Ex. Prove that pq  (p  q). F T T T F T T F F T T F T F T T F F F T

6. Equivalence Laws
propositional equivalences provide a pattern or template that can be used to match all or part of a much more complicated proposition and to find an equivalence for it.

7. Equivalence Laws - Examples
Identity: pT  p pF  p
Domination: pT  T pF  F
Idempotent: pp  p pp  p
Double negation: p  p
Commutative: pq  qp pq  qp
Associative: (pq)r  p(qr) (pq)r  p(qr)

8. More Equivalence Laws
Distributive: p(qr)  (pq)(pr) p(qr)  (pq)(pr)
De Morgan’s: (pq)  p  q (pq)  p  q
Trivial tautology/contradiction: p  p  T p  p  F

9. Defining Operators via Equivalences
Using equivalences, we can define operators in terms of other operators.
Exclusive or: pq  (pq)(pq) pq  (pq)(qp)
Implies: pq  p  q
Biconditional: pq  (pq)  (qp) pq  (pq)

10. An Example Problem
Check using a symbolic derivation whether (p  q)  (p  r)  p  q  r.
(p  q)  (p  r) 
[Expand definition of ] (p  q)  (p  r)
[Defn. of ]  (p  q)  ((p  r)  (p  r))
[DeMorgan’s Law]
 (p  q)  ((p  r)  (p  r))

11. Example Continued... (p  q)  ((p  r)  (p  r))  [ commutes]
 (q  p)  ((p  r)  (p  r)) [ associative]
 q  (p  ((p  r)  (p  r))) [distrib.  over ]
 q  (( (p  (p  r))  (p  (p  r)))
[assoc.]  q  (((p  p)  r)  (p  (p  r)))
[trivail taut.]  q  ((T  r)  (p  (p  r)))
[domination]  q  (T  (p  (p  r)))
[identity]  q  (p  (p  r))  cont.

12. End of Long Example q  (p  (p  r))
[DeMorgan’s]  q  (p  (p  r))
[Assoc.]  q  ((p  p)  r)
[Idempotent]  q  (p  r)
[Assoc.]  (q  p)  r
[Commut.]  p  q  r
Q.E.D. (quod erat demonstrandum)
(Which was to be shown.)

13. Another Example
Ex. 7) Show that (p  (p  q)) and p  q are logically equivalent by developing a series of logical equivalences.
Solution:
(p  (p  q))  p  ( p  q) De Morgan
 p  [( p)  q) De Morgan
 p  (p  q)  (p  p)  (p  q)
 F  (p  q)  (p  q)  F  p  q
Study Example 8.

14. Another Example cont. Exercise 29. Show that
(p  q)  (q  r)  (p  r ) is tautology
Solution:
? Use the truth table~!
Another solution (symbolic derivation):
(p  q)  (q  r)  (p  r )
 (p  q)  (q  r)  (p  r )
 (( p  q)  (q  r))  (p  r )

15. cont.  (( p  q)  (q  r))  (p  r )
 (p  q)  (p  r )  (q  r)
 ((p  q)  p)  (r  (q  r))
 (p  p)  (q  p)  (r  (q  r))
 T  (q  p)  (r  (q  r))

16. cont.  T  (q  p)  (r  (q  r))  (q  p)  (r  (q  r))
 (q  p)  (r  q)  (r  r))
 (q  p)  (r  q)  T)
 (q  p)  (r  q)
 q  q  p  r
 T  p  r  p  r  T  T
Study Odd Numbered Exercises~!
especially the simple ones

17. Predicate Logic
Predicate logic is an extension of propositional logic that permits concisely reasoning about whole classes of entities.
Propositional logic (recall) treats simple propositions (sentences) as atomic entities.
In contrast, predicate logic distinguishes the subject of a sentence from its predicate.
Remember these English grammar terms?

18. Subjects and Predicates
In the sentence “The dog is sleeping”:
The phrase “the dog” denotes the subject - the object or entity that the sentence is about.
The phrase “is sleeping” denotes the predicate- a property that is true of the subject.
In predicate logic, a predicate is modeled as a function P(·) from objects to propositions.
P(x) = “x is sleeping” (where x is any object).

19. More About Predicates
Convention: Lowercase variables x, y, z... denote objects/entities; uppercase variables P, Q, R… denote propositional functions (predicates).
Keep in mind that the result of applying a predicate P to an object x is the proposition P(x). But the predicate P itself (e.g. P=“is sleeping”) is not a proposition (not a complete sentence).
E.g. if P(x) = “x is a prime number”, P(3) is the proposition “3 is a prime number.”

20. Propositional Functions
A Statement of the form P(x1, x2, …,xn) is the value of the propositional function P at the n-tuple (x1, x2, …,xn), and P is also called a n-place predicate or a n-ary predicate
E.g. let P(x,y,z) = “x gave y the grade z”, then if x=“Mike”, y=“Mary”, z=“A”, then P(x,y,z) = “Mike gave Mary the grade A.”

21. Quantifiers Quantification Predicate Calculus
an important way of creating a proposition from propositional function.
Predicate Calculus
The area of logic that deals with predicates and quantifiers.

22. Universal Quantifier
The universal quantification of P(x) is the statement
“P(x) for all values of x in the domain.”
x P(x) denotes the universal quantification of P(x). Here  is called the universal quantifier.
The collection of values that a variable x can take is called x’s universe of discourse (domain of discourse/domain).

23. Universal Quantifier continued…
An element for which P(x) is false is called a conterexample of x P(x)
Ex. 9) Let Q(x) be the statement “x < 2.” What is the truth value of the quantification x Q(x) , where the domain consists of all real numbers?

24. Universal Quantifier continued…
Solution: Q(x) is not true for every real number x, because, for instance, Q(3) is false. That is, x=3 is a counterexample for the statement x Q(x). Thus,
x Q(x)
is false.

25. Existential Quantifier
The existential quantification of P(x) is the proposition
“There exists an element x in the domain such that P(x).”
 x P(x) denotes the existential quantification of P(x). Here  is called the existential quantifier.

26. Existential Quantifier continued…
Ex. 16) What is the truth value of x P(x) , where P(x) is the statement “x2 > 10” and the universe of discourse consists the positive integers not exceeding 4?
Solution: Because the domain is {1, 2, 3, 4}, the proposition x P(x) is the same as the disjunction P(1)  P(2)  P(3)  P(4). Because P(4), which is the statement “42 > 10,” is true, it follows that is x P(x) false.

27. Free and Bound Variables
An expression like P(x) is said to have a free variable x (meaning, x is undefined).
A quantifier (either  or ) operates on an expression having one or more free variables, and binds one or more of those variables, to produce an expression having one or more bound variables.

28. Example of Binding y x P(x,y) has 2 free variables, x and y.
x P(x,y) has 1 free variable, and one bound variable. [Which is which?]
“P(x), where x=3” is another way to bind x. : P(3)
An expression with zero free variables is a bona-fide (actual) proposition.
An expression with one or more free variables is still only a predicate: x P(x,y) y x

29. Still More Conventions
Sometimes the universe of discourse is restricted within the quantification, e.g.,
x>0 P(x) is shorthand for “For all x that are greater than zero, P(x).” =x (x>0  P(x))
x>0 P(x) is shorthand for “There is an x greater than zero such that P(x).” =x (x>0  P(x))

30. More to Know About Binding
(x P(x))  Q(x) - The variable x is outside of the scope of the x quantifier, and is therefore free. Not a proposition!
(x P(x))  (x Q(x)) – This is legal, because there are 2 different x’s!

31. Quantifier Equivalence Laws
Definitions of quantifiers: If u.d.=a,b,c,… x P(x)  P(a)  P(b)  P(c)  … x P(x)  P(a)  P(b)  P(c)  …
From those, we can prove the laws: x P(x)  x P(x) x P(x)  x P(x)
Which propositional equivalence laws can be used to prove this?
(Chalkboard.) Another way to see why the order of quantifiers matters is to expand out the definitions of FORALL and EXISTS in terms of AND and OR. For example, suppose the universe of discourse just consists of two objects a and b. Now, consider some predicate P(x,y).
Then,
FORALL x EXISTS y P(x,y)  (EXISTS y P(a,y)) /\ (EXISTS y P(b,y))
 (P(a,a) \/ P(a,b)) /\ P(b,a) \/ P(b,b)).
In contrast,
EXISTS y FORALL x P(x,y)  (FORALL x P(x,a)) \/ (FORALL x P(x,b))
 (P(a,a) /\ P(b,a)) \/ (P(a,b) /\ P(b,b)).
To see that these two are inequivalent, suppose only P(a,a) and P(b,b) are true. Then, the first proposition (with the FORALL first) is true, but, the second proposition (with the EXISTS first) is true. Students can come up with this counterexample in-class as an exercise.
DeMorgan's

32. More Equivalence Laws
x y P(x,y)  y x P(x,y) x y P(x,y)  y x P(x,y)
x (P(x)  Q(x))  (x P(x))  (x Q(x)) x (P(x)  Q(x))  (x P(x))  (x Q(x))

33. Negating Quantified Expressions
Consider the negation of the statement
“Every student in your class has taken a course in calculus.”
x Q(x), where Q(x) is “x has taken a course in calculus”, and the domain consists of the students in your class.
The negation is “There is a student in your class who has not taken a course in calculus.”
 x Q(x) ≡ x Q(x), x Q(x) ≡ x Q(x)

34. English to Logical Expressions
Ex. 23) Express the statement “Every student in this class has studied calculus” using predicates and quanifiers.
Solution: Let’s rewrite the sentence,
=> “For every student in this class, that student has studied calculus.”
=> “For every student x in this class, x has studied calculus.”

35. English to Logical Expressions cont.
=> x C(x)
(C(x): “x has studied calculus”, domain: student in the class)
=> “For every person x, if person x is a student in this class, x has studied calculus.”
=> x (S(x)  C(x))
(S(x): “person x is in this class”, domain: all people)
causion! How about “x (S(x)  C(x))”?

36. English to Logical Expressions cont.
Ex. 24) Express the statements “Some student in this class has visited Mexico” and “Every student in this class has visited either Canada or Mexico” using predicates and quanifiers.
Solution: Let’s rewrite the first sentence,
=> “There is a student in this class with the property that the student has visited Mexico.”
=> “There is a student x in this class having property that the x has visited Mexico.”

37. English to Logical Expressions cont.
=> x M(x)
(M(x): “x has visited calculus”, domain: students in the class)
=> “There is a person x having the property that x is a student in this class and x has visited Mexico.”
=>  x (S(x)  M(x))
(S(x): “person x is in this class”, domain: all people)
causion! How about “ x (S(x)  M(x))”?
Study the rest of this example!

38. Nesting of Quantifiers
Example: Let the u.d. of x & y be people.
Let L(x,y)=“x likes y” (a predicate w. 2 f.v.’s)
Then y L(x,y) = “There is someone whom x likes.” (A predicate w. 1 free variable, x)
Then x (y L(x,y)) = “Everyone has someone whom they like.” (A __________ with ___ free variables.)
Proposition

39. Quantifier Exercise
If R(x,y)=“x relies upon y,” express the following in unambiguous English:
x(y R(x,y))=
y(x R(x,y))=
x(y R(x,y))=
y(x R(x,y))=
x(y R(x,y))=
Everyone has someone to rely on.
There’s a poor overburdened soul whom everyone relies upon (including himself)!
There’s some needy person who relies upon everybody (including himself).
Everyone has someone who relies upon them.
Everyone relies upon everybody, (including themselves)!

40. Quantifier Exercise cont.
Ex. 4) Let Q(x, y) denote “x + y = 0.” What are the truth values of the quantifications xy Q(x,y) and xy Q(x,y), where the domain for all variables consists of all real numbers?
Solution: The quantifications xy Q(x,y) denotes
“There is a real number y such that for every real number x, Q(x, y).”
=> false. (See our textbook to study the reason! )

41. Quantifier Exercise cont.
The quantifications xy Q(x,y) denotes
“For every real number x there is a real number y such that Q(x, y).”
=> Given a real number x, there is a real number y such that x + y = 0; namely, y = -x. Hence, the statement xy Q(x,y) is true.
Study the explanation below this example (pp. 53).

42. Natural language is ambiguous!
“Everybody likes somebody.”
For everybody, there is somebody they like,
x y Likes(x,y)
or, there is somebody (a popular person) whom everyone likes?
y x Likes(x,y)
“Somebody likes everybody.”
Same problem: Depends on context, emphasis.