CSI Equivalences #40 Q6) (5 points) 1.4 Predicates and

CSI 2101AB, Winter 2018

Assignment 1 (5% of
final mark)

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Q1)                 
(5 points) 1.1 Propositional Logic
exercise #42

Q2)                 
(2 points) 1.1 Propositional Logic
exercise #46

Q3)                 
(3 points) 1.2 Applications of
Propositional Logic #12

Q4)                 
(2 points) 1.3 Propositional
Equivalences #36

Q5)                 
(1 point) 1.3 Propositional
Equivalences #40

Q6)                 
(5 points) 1.4 Predicates and
Quantifiers #32

Q7)                 
(4 points) 1.4 Predicates and
Quantifiers #42

Q8)                 
(2 points) 1.4 Predicates and
Quantifiers #46

Q9)                 
(5
points) 1.5 Nested Quantifiers #18

Q10)             
(1
point) 1.5 Nested Quantifiers #34

Q11)             
(4
points) 1.5 Nested Quantifiers #38

Answers:

1.      1.1
Propositional Logic exercise #42

a.     
If x=1, then (1+2=3) is true, so x:=x+1 is
executed. Final value of x is 2.

b.     
If x=1, then (x+1=3) is false and (2x+2=3) is
false, which makes ((x+1=3)OR(2x+2=3)) false, so x:=x+1 is not executed. Final
value of x is 1.

c.     
If x=1, then (2x+3=5) is true and (3x+4=7) is
true, which makes ((2x+3=5) AND (3x+4=7)) true, so x:=x+1 is executed. Final
value of x is 2.

d.     
If x=1, then (x+1=2) is true and (x+2=3) is true,
so ((x+1=2)XOR(x+2=3)) is false, so x:=x+1 is executed. Final value of x is 1.

e.     
If x=1, then (x<2) is true, so x:=x+1 is executed. Final value of x is 2.   2.       1.1 Propositional Logic exercise #46 "Fred is happy." – 0.8 "John is happy." – 0.4 "Fred and John are happy." – min(0.8, 0.4) = 0.4 "Neither Fred nor John are happy." – min(1-0.8, 1-0.4) = min(0.2, 0.6) = 0.2   3.      1.2 Applications of Propositional Logic #12 Let w denote "the file system is locked". Let x denote "new messages will be queued". Let y denote "the system is functioning normally, and conversely". Let z denote "the new messages will be sent to the message buffer".   Then these are the specifications written with the variables above:   "If the file system is not locked, then new messages will be queued." : ¬w à x   "If the file system is not locked, then the system is functioning normally, and conversely." : ¬w à y   "If new messages are not queued, then they will be sent to the message buffer." : ¬x à z   "If the file system is not locked, then the new messages will be sent to the message buffer." : ¬w à z   "New messages will not be sent to the message buffer." ¬z   The following configuration can allow the system to be consistent: w is true, x is true, y is false, and z is false. Specification Condition Truth Value ¬w à x F à T T ¬w à y F à F T ¬x à z F à F T ¬w à z F à F T ¬z T T       4.      1.3 Propositional Equivalences #36 s*=s when s=p?T because p?T is equivalent to p. Meanwhile s*=p?F which is also equivalent to p. Therefore, since p?T= p?F, s=s*.   5.      1.3 Propositional Equivalences #40 To make all parts of a conjunction proposition true, we can negate the false values. Therefore, the proposition would be p?q?¬r which is only true when p is true, q is true, and r is false.   6.      1.4 Predicates and Quantifiers #32 a.      Let x represent dogs and P(x) represent "x has fleas". So the given statement is ?xP(x), the negation is ?x¬P(x), and the negated expression is "There is a dog that does not have fleas". b.      Let x represent horses and P(x) represent "x can add". So the given statement is ?xP(x), the negation is ?x¬P(x), and the negated expression is "All horses cannot add". c.      Let x represent koalas and P(x) represent "x can climb". So the given statement is ?xP(x), the negation is ?x¬P(x), and the negated expression is "There is a koala that cannot climb". d.      Let x represent monkeys and P(x) represent "x can speak French". So the given statement is ?x¬P(x), the negation is ?xP(x), and the negated expression is "There is a monkey that can speak French". e.      Let x represent pigs and P(x) represent "x can swim" and Q(x) represent "x can catch fish". So the given statement is ?x(P(x)?Q(x)), the negation is ?x¬(P(x)?Q(x)), and the negated expression is "There is no pig that can swim and catch fish".   7.      1.4 Predicates and Quantifiers #42 a.      Let x represent users and P(x) represent "x has access to an electronic mailbox". Then the given statement becomes ?xP(x). b.      Let x represent the group members, y represent the file system, P(x) represent "x has access to an electronic mailbox", and Q(y) represent "y is locked". Then the given statement becomes Q(y)à?xP(x). c.      Let x represent the firewall, y represent the proxy server, and P(x) represent "x is in a diagnostic state". Then the given statement becomes P(x)<->P(y).

d.     
Let x represent routers, y represent the
throughput, z represent the proxy server, P(x) represent “x is functioning
normally”, Q(y) represent “y is between 100 kbps and 500 kbps”, and R(z)
represent “z is in diagnostic mode”. Then the given statement becomes (Q(y)?¬R(z))à?xP(x).

 

8.      1.4
Predicates and Quantifiers #46

 

 

9.      1.5
Nested Quantifiers #18

 

 

10.   1.5
Nested Quantifiers #34

 

 

11.   1.5
Nested Quantifiers #38