# CSC3190 Course Info by h7pc3hf

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```									          Lecture 3: Logic (2)

   Propositional Equivalences
   Predicates and Quantifiers

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3.1. Propositional Equivalences

   Categories of compound propositions:
• A tautology is a proposition which is always
true.
• Classic Example: PP
• A contradiction is a proposition which is
always false.
• Classic Example: PP
• A contingency is a proposition which neither a
• Example: (PQ)R
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3.1. Propositional Equivalences

   Two propositions P and Q are logically
equivalent if P  Q is a tautology. We
write
PQ

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3.1. Propositional Equivalences

   Example: (PQ)(QP)  (P  Q)
   Proof:
• Left side and the right side must have the
same truth values, independent of the
truth value of the component propositions.
• To show a proposition is not a tautology:
use an abbreviated truth table
• try to find a counter example or to disprove the
assertion.
• search for a case where the proposition is false.

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3.1. Propositional Equivalences
• Two possible cases:
P Q      PQ
•   Case 1: Left side false, right side true.
0   0          1
•   Case 2: Left side true, right side false.
0   1          1
1
1
0
1
0
1
• Case 1: Try left side false, right side true
•   Left side false: only one of PQ or QP need be false.
1a. Assume PQ = F. Then P = T, Q = F. But then
P   Q     PQ                  right side PQ = F. Oops, wrong guess.
0   0          1            1b. Try QP = F. Then Q = T, P = F. But then right
0   1          0
side PQ = F. Another wrong guess.
1   0          0
1   1          1
*Proof for (PQ)(QP)  (P  Q)

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3.1. Propositional Equivalences
• Case 2. Try left side true, right side false
P   Q     PQ
•   If right side is false, P and Q cannot have the same truth value.
0   0          1
2a. Assume P =T, Q = F. Then PQ = F and the conjunction
0   1          1
must be false so the left side cannot be true in this case.
1   0          0
Another wrong guess.
1   1          1
2b. Assume Q = T, P = F. Then QP = F. Again the left side
cannot be true.
P   Q     PQ      • We have exhausted all possibilities and not found a counter-
0   0          1       example. The two propositions must be logically equivalent.
0   1          0   •   Note: Given such equivalence, if and only if or iff is also stated
1   0          0       as is a necessary and sufficient condition for.
1   1          1

*Proof for (PQ)(QP)  (P  Q)

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3.1. Propositional Equivalences

   Some important logical equivalences:
Equivalences      Name
P T   PT

PT  P                                          0 1    1
Identity laws
PF  P
1 1    1

PT  T                                          P P   PP
Domination laws
PF  F
0 0    0
1 1    1

PP  P
Idempotent laws
PP  P
Double negation
(P)  P         law
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3.1. Propositional Equivalences

   Some important logical equivalences:
Equivalences                     Name
PQ  QP
Commutative laws
PQ  QP
(PQ)R  P(QR)
Associative laws
(PQ)R  P(QR)
P(QR)  (PQ)(PR)
Distributive laws
P(QR)  (PQ)(PR)
(PQ)  PQ
De Morgan’s laws
(PQ)  PQ
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3.1. Propositional Equivalences

   Other logical equivalences:
Equivalences                     Name
PQ  PQ                       Implication
PP  T                         Tautology
(PQ)(QP)  (PQ)              Equivalence
(PQ)(PQ)  P                Absurdity

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3.1. Propositional Equivalences

   Other logical equivalences:
Equivalences                     Name
(PQ)  (QP)                  Contrapositive
P(PQ)  P
Absorption
P (PQ)  P
(P  Q)R  P(QR)              Exportation
   Equivalent expressions can always be substituted
for each other in a more complex expression
– useful for simplification.
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3.1. Propositional Equivalences
   Example:
•   (P(PQ)) can be simplified by using the following
series of logical equivalence:
(P(PQ))
 P(PQ))         from the second De Morgan’s law
 P[(P)Q] from the first De Morgan’s law
 P(PQ)          from the double negation law
 (PP)(PQ] from the distributive law
 F(PQ)          since PPF
 PQ              from the identity law for F
 (PQ)             from the second De Morgan’s law

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3.1. Propositional Equivalences

REMEMBER!

We can always use a truth table to show that the simplified
proposition is equivalent to the original proposition.

   But Complexity (2n)…

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3.2. Predicates and Quantifiers

   2+1=3: a proposition
   x+y=3: propositional functions or predicates
•   A generalization of propositions
•   Propositions which contain variables
•   Predicates become propositions once every variable is
bound - by
• assigning it a value from the Universe of Discourse U
or
• quantifying it

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3.2.1. Predicates

   Example 1:
•   Let U = Z, the integers = , -2, -1, 0 , 1, 2, 3, 
•   P(x): x > 0, a predicate or propositional function.
•   It has no truth value until the variable x is bound.
•   Examples of propositions where x is assigned a value:
• P(-3) is false, i.e. -3 > 0 is false.
• P(0) is false.
• P(3) is true.
•   The collection of integers for which P(x) is true are the
positive integers.

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3.2.1. Predicates

   Example 1 (continued):
•   P(x): x > 0 is the predicate.
• P(y)P(0) is not a proposition. The variable y has not
been bound.
•   However, P(3)P(0) is a proposition which is true.
   Example 2:
•   Let R be the three-variable predicate R(x, y, z): x + y =
z. Find the truth value of
• R(2, -1, 5)
• R(3, 4, 7)
• R(x, 3, z)

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3.2.2. Quantifiers

   Specific value vs. Range
   What range of values in U for which the
bounded propositions are true?
   Two possibilities:
•   Universal: For all values in U
•   Existential: For some values in U

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3.2.2. Quantifiers
   Universal
•   P(x) is true for every x in the universe of discourse.
•   Notation: universal quantifier x P(x)
• For all x, P(x)
or
• For every x, P(x)
•   The variable x is bound by the universal quantifier
producing a proposition.
•   Example:
• U = { 1,2,3 }
x P(x)  P(1)P(2)P(3)

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3.2.2. Quantifiers
   Existential
•   P(x) is true for some x in the universe of discourse.
•   Notation: existential quantifier x P(x)
• There is an x such that P(x)
or
• For some x, P(x)
• For at least one x, P(x)
• I can find an x such that P(x)
•   Example:
• U = { 1,2,3 }
x P(x)  P(1)P(2)P(3)

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3.2.2. Quantifiers
REMEMBER!
A predicate (propositional function) is not a proposition until all variables
have been bound either by quantification or assignment of a value!

   Predicate equivalences:
• Equivalences involving the negation operator
• x P(x)  x P(x)
• x P(x)  x P(x)
• Distributing a negation operator across a
quantifier changes a universal to an
existential, and vice versa.

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3.2.2. Quantifiers

   Multiple Quantifiers:
•   Read from left to right . . .
•   Example 1:
• Let U = R, the real numbers,
P(x,y): xy = 0
xy P(x,y)
xy P(x,y)
xy P(x,y)
xy P(x,y)
•   The only one that is false is the first one. Why?
•   Suppose P(x,y) is the predicate x/y=1?

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3.2.2. Quantifiers

   Dangerous situations:
• Commutativity of quantifiers
xy P(x, y)  yx P(x, y)? YES!
xy P(x, y)  yx P(x, y)? NO!
DIFFERENT MEANING!

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3.2.2. Quantifiers

   Example of non-commutativity of quantifiers:
•   Let Q(x, y) denote “x + y = 0.”
•   Are the truth values of the quantifications yx P(x, y)
and xy P(x, y) the same?
   The answer is NO since:
•   yx P(x, y) means “There is a real number y such
that for all real numbers x, Q(x, y) is true.”
• The statement is false. Why?
•   x y P(x, y) means “For every real number x there is
a real number y such that Q(x, y) is true.”
• The statement is true.
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3.2.3. Converting from English

   (can be very difficult)
   Example 1:
• Express the statement “If somebody is female
and is a parent, then this person is someone’s
mother” as a logical expression.
• Let
F(x): x is female.
P(x): x is a parent
M(x, y): x is the mother of y.
• The statement applies to all people.
x ((F(x)  (P(x))  y M(x, y))

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3.2.3. Converting from English

   Example 2:
• Express the statement “Everyone has exactly
one best friend” as a logical expression.
• Let
B(x, y): y is the best friend of x.
• The statement says “exactly one best friend”. This
means that if y is the best friend of x, then all other
people z other than y can not be the best friend of
x.
xyz (B(x, y)  ((z  y)  B(x, z)))

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3.2.3. Converting from English

   Example 3:
• Consider the following statements.
“All lions are fierce.”
“Some lions do not drink coffee.”
“Some fierce creatures do not drink coffee.”
•   The first two are called premises and the third
is called the conclusion. The entire set is
called an argument.

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3.2.3. Converting from English

   Example 3 (continued):
• We can express these statements as follows.
• Let    P(x): x is a lion.
Q(x): x is fierce.
R(x): x drinks coffee.
• Then x (P(x)  Q(x)).
x (P(x)  R(x)).
x (Q(x)  R(x)).
• Why can’t we write the second statement as
x (P(x)  R(x))?

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