ICS 6D ( Discrete Mathematic): Solution to Homework #2
Section 1.5
4.
(a)
simplification
(b)
disjunctive syllogism
(c)
modus ponens
(d)
addition
(e)
hypothetical syllogism
6.
Let
p
be ”It rains”,
q
be ”It is foggy”,
r
be ”Sailing race is held”,
s
be ”Lifesaving demonstration goes
on” and
t
be ”Trophy is awarded”. Our premises are (
¬
p
∨ ¬
q
)
→
(
r
∧
s
),
r
→
t
and
¬
t
. And the conclusion
is
p
. The following argument shows how to derive the truth of
p
.
r
→
t
¬
t
:
¬
r
¬
r
:
¬
r
∨ ¬
s
(
¬
p
∨ ¬
q
)
→
(
r
∧
s
)
¬
(
r
∧
s
)
:
¬
(
¬
p
∨ ¬
q
)
p
∧
q
:
p
(Above we also used the DeMorgan Law twice: First that
¬
r
∨ ¬
s
is equivalent to
¬
(
r
∧
s
), and then the
same for
p, q
instead of
r, s
, which implies in particular that
¬
(
¬
p
∨ ¬
q
) is equivalent to
p
∧
q
.
24.
The error occurs in step 2. We cannot say that
p
(
c
) is true when
p
(
c
)
∨
q
(
c
) is true. In fact this might
look like simplification rule but note that simplification only applies for conjunction.
Section 2.1
4.
There are 4
×
4 different combination for subsets from which the following statements are true.
A
⊆
A
,
B
⊆
A
,
B
⊆
B
,
C
⊆
A
,
C
⊆
C
,
C
⊆
D
,
D
⊆
D
.
8.
(a)
True,
(b)
True,
(c)
False,
(d)
True,
(e)
True,
(f)
True,
(g)
True.
22.
(a)
It is not power set,
(b)
Power set of
{
a
}
,
(c)
It is not power set,
(d)
Power set of
{
a, b
}
34.
(a)
There exists a real number
x
, such that
x
3
=
−
1. ”True”
(b)
There exists an integer
x
, such that
x
+ 1
> x
. ”True”
(c)
For every integer
x
,
x
−
1 is an integer. ”True”
(d)
For every integer
x
,
x
2
is an integer. ”True”
Section 2.2
2.
(a)
A
∩
B
,
(b)
A
−
B
,
(c)
A
∪
B
,
(d)
¯
A
∪
¯
B
4.