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Chapter 8
Logic and Inference
8.2 Proposition Logic
8.3 Logical Inference
8.4 Predicate Logic
8.2 Proposition Logic
Proposition is a statement in terms of a variable which
should be True (1) or False (0).
Example
Ans.
Can the proposition be classified ?
(1) Primitive Proposition: Simplest one
(2) Compound Proposition: More complicated
Five connectives：NOT(), AND(), OR() imply(), and if and
only if()
e. g.
p Professor Chung is our Discrete math teacher.
q It rains.
r The ground is wet.
e. g.
qr
q  r
It rains and the ground is wet.
It doesn’t rain or the ground is wet.
Example
q
r
q
qr
qr
qr
q r
qr
0
0
1
0
0
1
1
1
0
1
1
0
1
1
1
0
1
0
0
0
1
0
0
0
1
1
0
1
1
1
1
1
qr
q  r
Example Prove by contradiction.
Ans.
Key: Prove “(p  q)” is false.
  p  q     p  q 
 p  q
Example Proof by contradiction example
Ans.
If it rains, then the ground is wet
pq
pq
p  q
p  q
p  q
-q means the ground is not wet, but p means it doesn’t
rain.

Example Simply
(1)  q   p  q    p
(2)   p  q     r  q  
Ans.
(1)
 q   p  q    p
  q  p     p  q   p 
  q  p    p  q 
  p  q 
Ans.
(2)
   p  q     r  q   (DeMorgan‘s law, 迪摩根律)
   p  q    r  q  (DeMorgan's law)
   p  q   q   r (Absorption law, 吸收律)
 qr
8.3 Logical Inference
Premises  Conclusion
Tautology
p1  p2  ...  pk  C
e. g.
p
pq
q
(1)
p   p  q   q   p   p  q   q
 p    p  q   q
 p   p   q   q
  p  p    p   q   q
 1   p   q   q
 1  p   1  q   q
 p   q  q
1
(2)
Example Is the following inference correct?
p   q  r 
s  q
t
p  t
 rs
Ans.
Prove by contradiction!
 p   q  r      s  q    t     p  t    r  s   0
 r  s  0
 r  1& s  0
  p   q  0    1  q    t    p  t   1
 t  0& p  0
From
 p   q  0    1& p  0
q0
But 1  q   1  0   0 
We complete the proof.
8.4 Predicate Logic
Predicate logic
EQUAL( x, y )
EQUAL(5,5)  1& EQUAL(5, 6)  0
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Quantifiers!
Universal: 
Existential: 
Example
P  x
 x  4  x  3  0
Q  x  EVEN  x 
x  P  x   Q  x   is true
x  4makesP  x   1& Q  x   1
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Example
“Everyone has only one strongest subject.”
Ans.
X:person
Y:subject
Z:the other subjects
P(x,y):x’s strongest subject is the course y

x y z P  x, y     z  y   P  x, z  

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Example
(1) There is a person in this city who works in City
Government but does not knows anyone else in
the city.
(2) Express the negation of (1) such that there is no
implication operator and no negation is to the left
of a quantifier.
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Ans.
(1)xy  C  x   F  x, y  


(2) xy  C  x  F  x, y    xy  C  x   F  x, y  
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Example
(1)x  A  x   B  y    (xA( x)  xB( x))
(2)xA( x)  xB( x)  x( A( x)  B( x))
Ans: (1) True!
key: if x is in global domain for two
predicates, then x is feasible to
be in local domain for each predicate.
(2) False! The reverse is not true.
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