Problem 3 (6 points) Let D denote the domain of all people. Alex is an individua
ID: 3890234 • Letter: P
Question
Problem 3 (6 points) Let D denote the domain of all people. Alex is an individual in D. You are given the following predicates ·D(x) : x is a doctor ·M(x) : x has a medical license K(x, y) : x knows y ·T(x, y) : x trusts y Translate each statement below into symbolic predicate logic. a. Every doctor has a medical license. b. Everybody knows a doctor whom they trust c. Alex doesn't know anyone with a medical license. d. Nobody trusts all doctors. Problem 4 (3 points) Given the following proposition in symbolic form (vxEZa yEZ(xysx+y)v(y 1)) (P2) Write the negation of proposition (P2) in the simplest form (with the negation moved as far to the right as possible). For example, for the proposition: (v x ZX x = x + 1 ) the negation would be: (BxEZ(x x+1). Problem 5 (7 points)Explanation / Answer
Assume,
=and
=or
~=Not
Problem 3
a)
x (D(x) -> M(x))
b)
x y ( K(x , y) T(x,y) D(y) )
c)
x y ( ~K(x , y) M(y) )
d)
~ ( x x ( T(x,y) D(y) ) )
Or
x x ( ~( T(x,y) D(y) ) )
Problem 4
The given proposition is ( x Z)( y Z) ( ( x.y x2 + y2 ) ( y1 ) )
Apply negation from left to right step by step as follows:
~(( x Z)( y Z) ( ( x.y x2 + y2 ) ( y1 ) ))
( x Z)~( y Z) ( ( x.y x2 + y2 ) ( y1 ) )
( x Z)( y Z)~ ( ( x.y x2 + y2 ) ( y1 ) )
( x Z)( y Z)~ ( ( x.y x2 + y2 ) ( y1 ) )
( x Z)( y Z) (~ ( x.y x2 + y2 ) ~( y1 ) )
( x Z)( y Z) ( ( x.y > x2 + y2 ) ~( y1 ) )
( x Z)( y Z) ( ( x.y > x2 + y2 ) ( y=1 ) )
Problem 5
a)
L: system is locked
Q: new messages will be queued
P: system is functioning normally
O: Old messages will be sent to message buffer
N: new messages will be sent to message buffer
b)
~L -> Q (S1)
~L <-> P (S2)
~Q -> O (S3)
~L -> N (S4)
~N (S5)
c)
The system is true, if all above five specifications are true simultaneously.
This can be proved by simple reasoning as follows:
So all the specifications are true for the following assignments
N
False
False
False
L
True
True
True
P
False
False
False
Q
True
False
True
O
False
True
True
All system specifications are
True
True
True
Therefore the system is consistent
Proof using truth table:
A determine whether the system is consistent or not , a truth table can be constructed. For some assignment, if all five have truth values, then the system is consistent.
L
Q
P
O
N
~L
~Q
~L -> Q
~L <-> P
~Q -> O
~L -> N
~N
F
F
F
F
F
T
T
F
F
F
F
T
F
F
F
F
T
T
T
F
F
F
T
F
F
F
F
T
F
T
T
F
F
T
F
T
F
F
F
T
T
T
T
F
F
T
T
F
F
F
T
F
F
T
T
F
T
F
F
T
F
F
T
F
T
T
T
F
T
F
T
F
F
F
T
T
F
T
T
F
T
T
F
T
F
F
T
T
T
T
T
F
T
T
T
F
F
T
F
F
F
T
F
T
F
T
F
T
F
T
F
F
T
T
F
T
F
T
T
F
F
T
F
T
F
T
F
T
F
T
F
T
F
T
F
T
T
T
F
T
F
T
T
F
F
T
T
F
F
T
F
T
T
T
F
T
F
T
T
F
T
T
F
T
T
T
T
F
F
T
T
T
F
T
F
T
T
T
F
T
F
T
T
T
T
T
F
T
T
T
T
F
T
F
F
F
F
F
T
T
T
F
T
T
T
F
F
F
T
F
T
T
T
F
T
F
T
F
F
T
F
F
T
T
T
T
T
T
T
F
F
T
T
F
T
T
T
T
T
F
T
F
T
F
F
F
T
T
F
F
T
T
T
F
T
F
T
F
T
T
F
F
T
F
T
F
T
T
F
F
T
T
F
T
T
T
T
F
T
T
T
F
T
T
F
T
T
F
T
T
F
F
F
F
F
T
T
T
T
T
T
T
F
F
T
F
F
T
T
T
T
F
T
T
F
T
F
F
F
T
T
T
T
T
T
T
F
T
T
F
F
T
T
T
T
F
T
T
T
F
F
F
F
T
F
T
T
T
T
T
T
F
T
F
F
T
F
T
T
F
T
T
T
T
F
F
F
T
F
T
T
T
T
T
T
T
T
F
F
T
F
T
T
F
The bolded rows represents that all the specifications are true for 3 assignments.
Thus the system is consistent.
N
False
False
False
L
True
True
True
P
False
False
False
Q
True
False
True
O
False
True
True
All system specifications are
True
True
True
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