1
Propositional Logic
Part A
1.
G⊃W
2.
3.
(W · R)  G
T ⊃ (M ⊃ (B ⊃ E)
4.
5.
6.
(E ⊃ H) ⊃ J
C v (U ⊃ P)
~P · ~B / ~(P v B)
Part B
1. F
2. F
3. T
Part C
1. Self-contradiction
2. Tautology
3. Contingent proposition
Part D
1. Inconsistent and contradictory
2
2. Consistent and equivalent
3. Consistent, but neither equivalent nor contradictory
Part E
Translate and determine whether the following argument is valid with a truth table.
If high school graduates are deficient in reading, they will not be able to compete in the modern world. If high school graduates are deficient in writing, they
will not be able to compete in the modern world. Therefore, if high school graduates are deficient in reading, then they are deficient in writing.
P1. P  ~Q
P2. R  ~Q
C. P  R
T T
T T
 ~
F F
T T
F
F
T F
T T
T T F
T
T T F
T
T T
T F
F
T T F
F
T T F
T
F
F T
T F
T F
T
T
F
F
T
F
T T
F T
F
T F
T
F
T F
T
F
T F
F F
T F
T T F
T
T T F
F
T T
F F
F
T T F
F
T T F
F
T F
P Q R P
T
T
F
F
Q //
P
T
 ~
F F
T
T
 R
T T
T
F
T F
T
T
F
Q
T
F
/
R
F
F
The argument is invalid as it is logically possible that all premises are true and the conclusion is false (i.e. row 4).
3
Part F
Use truth table to determine whether the following arguments are valid.
(1) ~(K  S)
S  ~(R v K)
R  ~S
T
T T F
(K 
T T
T
T F
F
T
T
T
T
F
F
F
T
T
F
F F
T
T
F T T
T
F
F
F
T F
T
T
T
T
T T
F
T
F F
T
T
F
F
F
T F
F
T
T
F
T T
F
F
T T T
F
F
T
T
F
F
T
T
F
T
T F
T
F
T F
T
F
F
T
T
T T F
F
F
F
F F
T
F
F T F
F
T
F
F
T F
T
T
F
T
T T
F
F
F F
F
T
F
F
T T F
F
F
F
T T
F
K S R ~
F
S)
T
T
 ~
F F
/
S
(R 
T T
K)
R

~
S
T
T
T F
T
//
The argument is invalid as it is logically possible that all premises are true and the conclusion is false (i.e. row 6).
4
(2) G  H
RG
~H  G
RH
G H R G

H
R

G
~
H

G
T
T
T T
T
T
T
T
T
F
T
T
T
T
F
T
T
T
F
F
T
F
T
T
F
T T
F
F
T
T
T
T
T
F
F
T
F
F
F
F
T
F
T
T F
T
T
T
F
F
T
F
F
T
T
F
F
F
T F
T
F
F
F
F
T
F
F
/
R

H
T
T
T
T
T
T
F
F
T
F
T
T
T
F
F
T
F
T
T
F
T
F
F
F
T
F
F
T
T
T
T
F
F
T
F
F
F
F
T
T
F
F
T
F
T
F
T
F
F
F
T
F
T
F
T
F
F
T
F
/
//
The argument is valid as there is no logical possibility that all premises are true and the conclusion is false.
Part G
Use indirect truth table to determine whether the following arguments are valid.
(1)
First of all, we assume the argument as invalid by assign truth to its premises and falsity to its conclusion.
(P
T

T
(Q

T
R))
/
((P
T

R)

T
~
(S

T))
/
((~
S

~
T)

T
~(
P
T

Q))
//
(S
≡
F
T)
5
There is no way to continue filling the boxes without introducing an assumption.
Assumption: S is true
(P
T

T

T
(Q
T
R))
/
F
((P
T

F
R)
F

T
~
(S
F
T

T
T))
/
F
((~
S
F
T

F
~
T)
F
T

T

T
Q))
Q))
T

F
(K

J)
~(
P
F
T
~(
P
T
//
T
(S
≡
T)
T
F
F
(S
≡
T)
F
F
T
Given that there is a contradiction to our hypothesis that S is true, it implies that it is not the case that S is true.
Therefore, S is false.
(P
T

T

T
(Q
F
R))
/
T
((P
T

T
R)
T

T
~
(S
T
F

F
T))
/
F
((~
S
T
F

T
~
T)
F
T

T
//
F
Given that there is a contradiction again, we conclude that our initial assumption [that the argument is invalid] is false.
Therefore, the argument is valid.
(2)
(~
G

(~
H

~
I))
/
(J
T

H)
/
(K
T

(L

M))
/
T
//
(L
T

G)
F
Assumption: L is true
(~
G
T
F

T
(~
H
T
F

T
~
I))
T
F
/
(J
F

T
H)
F
/
(K
T

T
(L
T

T
M))
T
/
(K
T

T
J)
F
//
(L
T
There is no contradiction in the filling process. Therefore, the argument is invalid and the counter-example is {G: F, H:F, I:F, J:F, K:T, L:T, M:T}.1
1
You can assign the values states in the counter-example to the argument in order to reconfirm that the counter-example is correct.

F
G)
F
6
Part H
Use indirect truth table to determine whether the sets of propositions are valid.
(1)
First of all, we assume that all the formulas are consistent.
((G

T
~
Q)
(F
F

T
T
5
4
B))
F

T
3
1
3
6
/
Q)
F

F
F
3
2
3

T
G)
~(
F
T
T
7
1
/
N)
T

T
T
7
1
8
(B
/
N)
F

T
3
9
((F
Q)
T

T
8
1
10
Y

T
T
A contradiction arises from the assumption that they are consistent.
Therefore, they are inconsistent.
(2)
(G

T
P)
/
(P
1

T
(A

~
G))
/
((R

T)
1
/
Y
1

T
R
/
B
1

T
T
/
1
B
1
Assumption: G is true
P)
T

T
T
T

T
A
1
2
2
1
(G
/
(P
(A

T
~
G))
G)
T

T
T

T
F
3
4
A
1
A
1
A contradiction arises from the assumption that G is true.
Thus, G is false.
/
((R

T)
/
Y
R
/
B

T
1
T
/
Y

T
1
B
7
F

T
2
1
(G
P)
/
(P

T
1
(A

~
G))
T
F
3
2
/
T)
F

F
4
3
((R
G)
F

T
F
4
1
2
Contradiction arises and we can conclude that these formulas are inconsistent.
/
R
F

T
F
5
1
4
Y
/
T
F

T
F
5
1
4
B
/
B
F

T
5
1
6
Y
T