Solutions to Additional Problem Set 2

1
.
​
Out[]=
p
q
r
¬p(q∧r)
True
True
True
True
True
True
False
True
True
False
True
True
True
False
False
True
False
True
True
True
False
True
False
False
False
False
True
False
False
False
False
False
2
.
As we can see from the truth table below they are both equivalent.
Out[]=
p
q
r
p(q∨r)
(pq)∨(pr)
True
True
True
True
True
True
True
False
True
True
True
False
True
True
True
True
False
False
False
False
False
True
True
True
True
False
True
False
True
True
False
False
True
True
True
False
False
False
True
True
3
.
​
3
.
1
.
Start with
¬((¬P∧Q)∨¬(R∨¬S))
using DeMorgan with the outside negation we get
(¬(¬P∧Q)∧(R∨¬S))
Using DeMorgan with the negation outside of (¬P∧Q) we get
((P∨¬Q)∧(R∨¬S))
3
.
2
.
Start with
¬((¬P→¬Q)∧(¬Q→R))
Using that
AB
is equivalent to
¬A∨B
and applying it to both implications we get
¬((P∨¬Q)∧(Q∨R))
Using DeMorgan with the outermost negation we get
(¬(P∨¬Q)∨¬(Q∨R))
Using DeMorgan again
((¬P∧Q)∨(¬Q∧¬R))
3
.
3
.
Consider the following truth table
Out[]=
P
Q
R
S
¬((¬P∧Q)∨¬(R∨¬S))
((P∨¬Q)∧(R∨¬S))
¬((¬P→¬Q)∧(¬Q→R))
((¬P∧Q)∨(¬Q∧¬R))
True
True
True
True
True
True
False
False
True
True
True
True
True
True
False
False
True
True
False
False
False
False
False
False
True
True
False
True
True
True
False
False
True
False
True
True
True
True
False
False
True
False
True
True
True
True
False
False
True
False
False
False
False
False
True
True
True
False
False
True
True
True
True
True
False
True
True
False
False
False
True
True
False
True
True
False
False
False
True
True
False
True
False
False
False
False
True
True
False
True
False
False
False
False
True
True
False
False
True
True
True
True
False
False
False
False
True
True
True
True
False
False
False
False
False
False
False
False
True
True
False
False
False
True
True
True
True
True
​
4
.
​
4
.
1
.
Truth Table
Out[]=
T
S
P
(T∨S)¬P
True
True
True
False
True
True
False
True
True
False
True
False
True
False
False
True
False
True
True
False
False
True
False
True
False
False
True
True
False
False
False
True
4
.
2
.
A counter example would be a number that is a Prime, and also either a Triangular number or a Square.
4
.
3
.
If the implication were true (which it is not by the way) then we would have that 5657 is neither a square nor a triangular number.
5
.
P = Tommy had popcorn
R = Tommy had raisins
Q = Tommy had cucumber sandwiches
S = Tommy had a soda
T = Tommy had tea
​
Tommy said:
P∨R
QS
¬S∧¬T
​
The negations of those statements are:
¬P∧¬R
¬S∧Q
S∨T
​
Out[]=
P
Q
R
S
T
(¬P∧¬R)∧(¬S∧Q)∧(S∨T)
True
True
True
True
True
False
True
True
True
True
False
False
True
True
True
False
True
False
True
True
True
False
False
False
True
False
True
True
True
False
True
False
True
True
False
False
True
False
True
False
True
False
True
False
True
False
False
False
True
True
False
True
True
False
True
True
False
True
False
False
True
True
False
False
True
False
True
True
False
False
False
False
True
False
False
True
True
False
True
False
False
True
False
False
True
False
False
False
True
False
True
False
False
False
False
False
False
True
True
True
True
False
False
True
True
True
False
False
False
True
True
False
True
False
False
True
True
False
False
False
False
False
True
True
True
False
False
False
True
True
False
False
False
False
True
False
True
False
False
False
True
False
False
False
False
True
False
True
True
False
False
True
False
True
False
False
False
True
False
False
True
True
False
True
False
False
False
False
False
False
False
True
True
False
False
False
False
True
False
False
False
False
False
False
True
False
False
False
False
False
False
False
​
So when all the actually valid propositions are true only Q and T are true, that is : Tommy had cucumber sandwiches and tea.
6
.
For each of the following claims, determine (1) the minimum number of cards you must turn over to check the claim, and (2) what those cards are, in order to determine if the claim is true of all four cards.
6
.
1
.
If there is a P or Q on the letter side of the card, then there is a diamond on the shape side of the card.
Let:
PQ = There is a P or Q on the letter side of the card.
D = There is a diamond on the shape side of the card.
​
We need to check (PQ)D for each card
Out[]=
PQ
D
PQD
True
True
True
True
False
False
False
True
True
False
False
True
Card 1:
D is false so we are in lines 2 or 4 so we still need to check PQ
So needs to be flipped
​
Card 2:
D is false so we are in lines 2 or 4 so we still need to check PQ
So needs to be flipped
​
Card 3:
PQ is true, so we are in lines 1 or 2, so we still need to check D
So needs to be flipped
​
Card 4,
D is true so we are in lines 1 or 3, both giving PQD the truth value True
So it does not need to be flipped.
6
.
1
.
1
.
Three cards
6
.
1
.
2
.
Card 3, card 1, and card 2
6
.
2
.
If there is a Q on the letter side of the card, then there is either a diamond or a star on the shape side of the card.
Let
Q = There is a Q on the letter side of the card
DS = There is a diamond or star on the shape side of the card.
Out[]=
Q
DS
QDS
True
True
True
True
False
False
False
True
True
False
False
True
Card 1:
DS is True, so we are in lines 1 or 3, both giving QDS True.
So it does not need to be flipped.
​
Card 2:
DS is False, so we are in lines 2 or 4, depending on the value of Q.
It does need to be flipped
​
Card 3:
Q is True, so we are on lines 1 or 2, depending on the value of DS.
I does need to be flipped
​
Card 4:
DS is True, so we are in lines 1 or 3, both giving QDS True.
So it does not need to be flipped.
6
.
2
.
1
.
2 cards
6
.
2
.
2
.
Card 2 and card 3
7
.
KC = “Kronecker killed Cardinality”
BC = “Berkeley killed Cardinality”
KK = “Kronecker was in the kitchen”
BD = “Berkeley was in the drawing room”
BB = “Berkeley was wearing boots”
​
We have:
a. BC ∨ KC
b. KCKK
c. BCBD
d. BDBB
e. ¬BB
​
Look at the Truth tables for these statements:
Out[]=
0
BC ∨ KC
KCKK
BCBD
BDBB
¬BB
KC
1
True
True
True
True
False
True
2
True
True
True
False
True
True
3
True
True
False
True
False
True
4
True
True
False
True
True
True
5
True
False
True
True
False
True
6
True
False
True
False
True
True
7
True
False
False
True
False
True
8
True
False
False
True
True
True
9
True
True
True
True
False
True
10
True
True
True
False
True
True
11
True
True
True
True
False
True
12
True
True
True
True
True
True
13
True
False
True
True
False
True
14
True
False
True
False
True
True
15
True
False
True
True
False
True
16
True
False
True
True
True
True
17
True
True
True
True
False
False
18
True
True
True
False
True
False
19
True
True
False
True
False
False
20
True
True
False
True
True
False
21
True
True
True
True
False
False
22
True
True
True
False
True
False
23
True
True
False
True
False
False
24
True
True
False
True
True
False
25
False
True
True
True
False
False
26
False
True
True
False
True
False
27
False
True
True
True
False
False
28
False
True
True
True
True
False
29
False
True
True
True
False
False
30
False
True
True
False
True
False
31
False
True
True
True
False
False
32
False
True
True
True
True
False
Only line 12 has all premises true, and the conclusion KC is true in that case, so the argument is valid
8
.
​
8
.
1
.
​
Only in line 1 are all premises true, and in that case R is true, so the argument is valid
8
.
2
.
Consider the truth table
The premise shows the conclusion, moreover they are equivalent
8
.
3
.
Only two atomic propositions so we go with a truth table
Since we only care at the rows with P and ¬Q true (Row 2) we see that the truth value of ¬(P⧦Q)is true, then we conclude.
8
.
4
.
​
Similarly we care only about rows 1 and 4, so the conclusion is valid
8
.
5
.
We build a table in which just the premises are true
8
.
6
.
(P∨Q)⟺R)
¬P⟺Q
​
Both lines 3 and 5 have the true value for our premises, and in both cases the truth value for R is True, so the conclusion is valid
8
.
7
.
Notice by the table below that the conclusion is always true.
8
.
8
.
We proceed as in the exercise above.