Download Discrete mathematics- Math150

CSTS-SEU-KSA
Discrete mathematics- Math150
Assignment 1
2nd semester 2016-2017
------------------------------------------------------------------------------------------Section I
Determine whether the statement is true or false. (1 mark for each)
1) x + y = z is a proposition.
2) The compound proposition p ∧ p is a tautology.
3)
There are 256 different Boolean functions of degree 3.
4) If
5)
(
, then ̅̅̅̅̅̅̅̅̅̅̅̅
(
)=
)
̅̅ ̅
The argument in propositional logic is valid if the premises imply the
conclusion.
6) The expression “ (¬ p∧ q) ∨ r” is in disjunctive normal form (DNF)
Q
1
f
2
f
3
T
4
F
1
5
T
6
T
CSTS-SEU-KSA
Section II
Choose the correct answer
1) The converse of the statement q →r
(1 mark for each)
is
a) q →r
b) ¬ q →¬r
c) r →q
d) ¬ r →q
2) If P: It is raining , and Q: I will study discrete mathematics, then the
statement” if It is raining, then I will study discrete mathematics”
represents
P∨q
P∧q
Pq
Pq
a)
b)
c)
d)
3) Boolean expression for the Boolean function F(x, y) which defined by the
table below is
0
0
1
1
a)
b)
c)
d)
̅̅
̅
̅
̅̅̅
F( , )
1
1
1
0
0
1
0
1
̅
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4) The representation of the sentence "Every student in this university took a
course in mathematics," in predicate logic, is ----------a) xMx
b) xMx
c) x ¬ Mx
d) x ¬ Mx
5) Assume that p ↔q is True, then
a)
b)
c)
d)
Both of p and q must be same , otherwise false
Both of p and q must be true , otherwise false
No one of p and q must be true , otherwise false
Both of p and q must be false , otherwise false
6) Let Mx,y denoted by “
”,where
, then -----------is True.
a) xyMx
b) y xMx
c) y x Mx
d) xy ¬ Mx
1
C
2
C
3
A
4
A
5
A
3
6
B
CSTS-SEU-KSA
Section III
Solve the following questions
(3 marks for each)
1) Construct a truth table for the proposition
(¬ p → q) ∧ (q → r) → r.
P
T
T
F
F
F
T
F
T
q
T
F
T
F
T
F
F
T
r
F
T
T
T
F
F
F
T
¬p→q
T
T
T
F
T
T
F
T
q→r
F
T
T
T
F
T
T
T
(¬p → q) ∧ (q → r)
F
T
T
F
F
T
F
T
(¬ p → q) ∧ (q → r) → (r)
T
T
T
T
T
F
T
T
2) Show that p ↓q is logically equivalent to ¬ p∨ q) by using a truth table
p
T
T
F
F
q
T
F
T
F
p ↓q
F
F
F
T
¬ p∨ q)
F
F
F
T
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CSTS-SEU-KSA
3) Find the output of the following circuit
P
q
r
T
T
F
F
F
T
F
T
T
F
T
F
T
F
F
T
F
T
T
T
F
F
F
T

r
T
F
F
F
T
T
T
F
 r ∨ q  r ∨ q) ∧ P
T
F
T
F
T
T
T
T
T
F
F
F
F
T
F
T
4) Find the sum-of-products expansions of the Boolean
function F ( , ) = , by using a truth table.
x
0
1
1
0
y
1
0
1
0
o
o
1
o
F( , )=
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CSTS-SEU-KSA
5) What is the truth value of ∃xP(x) and  xP(x), where P(x) is the statement
“x2 > 5” and the domain is {1, 2, 3}?
∃xP(x)= P(1) ∨ P(2) ∨ P(3)=T
 xP(x)= P(1) ∧ P(2) ∧ P(3)=F
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CSTS-SEU-KSA
6) Find the Disjunctive Normal Form (DNF) of
((p∨q) ∨ (s∨r) ) → ¬ (p∨q)
SOLUTION
(¬p∧¬q) ∨ ¬((p∨q) ∨ (s∨r))
Construct the truth table for the proposition. Then an equivalent proposition is
the disjunction with n disjoints (where n is the number of rows for which the
formula evaluates to T). Note that, we used this idea in chapter 12.
p
q
r
s
¬p∧¬q
p∨q
s∨r
(p∨q) ∨
(s∨r)
¬((p∨q) ∨
(s∨r))
T
T
F
F
F
T
F
T
T
T
F
F
F
T
F
T
T
F
T
F
T
F
F
T
T
F
T
F
T
F
F
T
F
T
T
T
F
F
F
T
F
T
T
T
F
F
F
T
F
T
T
T
F
F
F
T
T
F
F
F
T
T
T
F
f
f
f
T
f
f
T
f
f
f
f
T
f
f
T
f
t
t
t
F
t
t
F
t
t
t
t
F
t
t
F
t
F
t
t
t
F
F
F
t
t
t
t
t
t
t
t
t
T
T
T
T
T
T
f
T
T
T
T
T
T
T
T
T
f
f
f
f
f
f
T
f
f
f
f
f
f
f
f
f
((p∨q) ∨ (s∨r) )
→ ¬ (p∨q)
F
F
F
T
F
F
T
F
F
F
F
T
F
F
T
F
(¬p∧¬q∧r∧s) ∨ (¬p∧¬q∧¬r∧¬s) ∨ (¬p∧¬q∧r∧¬s) ∨ (¬p∧¬q∧¬r∧s)
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