Wolfram Cloud Document

Solutions to Additional Problem Set 2

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

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

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))

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))

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

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

A counter example would be a number that is a Prime, and also either a Triangular number or a Square.

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.

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.

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.

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.

Three cards

Card 3, card 1, and card 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.

2 cards

Card 2 and card 3

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

Only in line 1 are all premises true, and in that case R is true, so the argument is valid

Consider the truth table

The premise shows the conclusion, moreover they are equivalent

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.

Similarly we care only about rows 1 and 4, so the conclusion is valid

We build a table in which just the premises are true

(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

Notice by the table below that the conclusion is always true.

We proceed as in the exercise above.