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Mathematics for Comter I 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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2006 310205 Mathematics for Comter I 1
Lecture 3: Logic (2) Propositional Equivalences Predicates and Quantifiers
2006 310205 Mathematics for Comter I 2
3.1. Propositional Equivalences Categories of compound propositions:
• A tautology is a proposition which is always true.• Classic Example: PP
• A contradiction is a proposition which is always false.• Classic Example: PP
• A contingency is a proposition which neither a tautology nor a contradiction.• Example: (PQ)R
2006 310205 Mathematics for Comter I 3
3.1. Propositional Equivalences Two propositions P and Q are logically
equivalent if P Q is a tautology. We write
P Q
2006 310205 Mathematics for Comter I 4
3.1. Propositional Equivalences Example: (PQ)(QP) (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:
• Case 1: Left side false, right side true.• Case 2: Left side true, right side false.
• Case 1: Try left side false, right side true• Left side false: only one of PQ or QP need be false.
1a. Assume PQ = F. Then P = T, Q = F. But then right side PQ = F. Oops, wrong guess.
1b. Try QP = F. Then Q = T, P = F. But then right side PQ = F. Another wrong guess.
*Proof for (PQ)(QP) (P Q)
P Q PQ
0 0 1
0 1 1
1 0 0
1 1 1
P Q PQ
0 0 1
0 1 0
1 0 0
1 1 1
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3.1. Propositional Equivalences• Case 2. Try left side true, right side false
• If right side is false, P and Q cannot have the same truth value.2a. Assume P =T, Q = F. Then PQ = F and the conjunction
must be false so the left side cannot be true in this case. Another wrong guess.
2b. Assume Q = T, P = F. Then QP = F. Again the left side cannot be true.
• We have exhausted all possibilities and not found a counter-example. The two propositions must be logically equivalent.
• Note: Given such equivalence, if and only if or iff is also stated as is a necessary and sufficient condition for.
*Proof for (PQ)(QP) (P Q)
P Q PQ
0 0 1
0 1 1
1 0 0
1 1 1
P Q PQ
0 0 1
0 1 0
1 0 0
1 1 1
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3.1. Propositional Equivalences Some important logical equivalences:
Equivalences Name
PT P
PF PIdentity laws
PT T
PF FDomination laws
PP P
PP PIdempotent laws
(P) PDouble negation law
P T PT
0 1 1
1 1 1
P P PP
0 0 0
1 1 1
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3.1. Propositional Equivalences Some important logical equivalences:
Equivalences Name
PQ QP
PQ QPCommutative laws
(PQ)R P(QR)
(PQ)R P(QR)Associative laws
P(QR) (PQ)(PR)
P(QR) (PQ)(PR)Distributive laws
(PQ) PQ
(PQ) PQDe Morgan’s laws
2006 310205 Mathematics for Comter I 9
3.1. Propositional Equivalences Other logical equivalences:
Equivalences Name
PQ PQ Implication
PP T Tautology
PP F Contradiction
(PQ)(QP) (PQ) Equivalence
(PQ)(PQ) P Absurdity
2006 310205 Mathematics for Comter I 10
3.1. Propositional Equivalences Other logical equivalences:
Equivalent expressions can always be substituted for each other in a more complex expression – useful for simplification.
Equivalences Name
(PQ) (QP) Contrapositive
P(PQ) P
P (PQ) PAbsorption
(P Q)R P(QR) Exportation
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3.1. Propositional Equivalences Example:
(P(PQ)) can be simplified by using the following series of logical equivalence:
(P(PQ)) P(PQ)) from the second De Morgan’s law P[(P)Q] from the first De Morgan’s law P(PQ) from the double negation law (PP)(PQ] from the distributive law F(PQ) since PPF PQ from the identity law for F (PQ) from the second De Morgan’s law
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3.1. Propositional Equivalences
But Complexity (2n)…
REMEMBER!
We can always use a truth table to show that the simplified proposition is equivalent to the original proposition.
2006 310205 Mathematics for Comter I 13
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 Uor• 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
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.
REMEMBER!
A predicate (propositional function) is not a proposition until all variables have been
bound either by quantification or assignment of a value!
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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
xy P(x,y)xy P(x,y)xy P(x,y)xy 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 quantifiersxy P(x, y) yx P(x, y)? YES!
xy P(x, y) yx 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 yx P(x, y)
and xy P(x, y) the same? The answer is NO since:
yx 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 parentM(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.
xyz (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 asx (P(x) R(x))?
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3.3. Further Readings Propositional Equivalences
• Rosen: Section 1.2. Predicates and Quantifiers
• Rosen: Section 1.3.