Please help answer these questions, thank you. (3 marks) How many rows appear in
ID: 3877077 • Letter: P
Question
Please help answer these questions, thank you.
(3 marks) How many rows appear in the truth table for each of the following compound propositions? All lowercase letters are propositions. 3. (10 marks) Construct truth tables for each of the following compoun a column for all statements used in the compound proposition, ie -q should be a column in 4. d propositions. Include (a) (p q ) v ( p ^q) 5. (10 marks) Determine whether the following statements are valid. Demonstrate how you derived the result. (a) ( p ^ q ) (p V q ) is a tautology. ( p q) ^ q ^ p) is contradiction 6. (10 marks) Using the laws of propositional logic, determine whether the following statements are logically equivalent:Explanation / Answer
3)
Generally the number of rows in the truth table for n number of propositions is 2^n
a) 4 rows are appear in the truth table because it consist of 2 propositions
b) 16 rows are appear in the truth table because it consist of 4 propositions
c) 8 rows are appear in the truth table because it consist of 3 propositions
4)
a) (p q) (p q)
p
q
(p q)
(p q)
((p q) (p q))
0
0
1
0
1
0
1
0
0
0
1
0
0
0
0
1
1
1
1
1
b) (qr)(¬qr)
q
r
¬q
(q r)
(¬q r)
((q r) (¬q r))
0
0
1
0
1
0
0
1
1
0
1
0
1
0
0
0
0
1
1
1
0
1
1
1
5)
Tautology - if all the final values in the truth table are result to true then we say that given statement is tautology.
Contradiction - if all the final values in the truth table are result to false then we say that given statement is contradiction.
a) ((p q) (p q))
p
q
(p q)
(p q)
((p q) (p q))
0
0
0
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
1
For the given statement the resultant values all are true so given statement is tautology.
b) ((p q) ¬(¬q ¬p))
p
q
¬p
¬q
pq
¬q¬p
¬(¬q¬p)
(pq)¬(¬q¬p)
0
0
1
1
1
1
0
0
0
1
1
0
0
0
1
0
1
0
0
1
0
0
1
0
1
1
0
0
1
0
1
1
For the given statements the resultant values are not all false so the given statement is not a contradiction.
6)
a) ((p q) (p r)) = (p (q r))
p
q
r
((p q) (p r))
(p (q r))
0
0
0
1
1
0
0
1
1
1
0
1
0
1
1
0
1
1
1
1
1
0
0
0
0
1
0
1
1
1
1
1
0
1
1
1
1
1
1
1
When we observe the result of LHS and RHS of given equivalence statement they both are same so given statement are logically equivalent.
By using propositional Laws:
((p q) (p r)) = (p (q r))
= (¬pq)(¬pr) pq = ¬pq based on the propositonal equvalance laws
= ¬p(qr) take ¬p as common
= ¬(¬p)(qr) pq = ¬pq so assume ¬p as p and qr as q
= p(qr) ¬(¬p) = p
Therefore
LHS = RHS so both are equivalent.
b) ((p q) (p q))
P
q
(p q)
(pq)
((p q) (p q))
T
0
0
0
1
1
1
0
1
0
1
1
1
1
0
0
0
1
1
1
1
1
1
1
1
When we observe the result of LHS and RHS of given equivalence statement they both are same so given statement are logically equivalent.
By using Propositional Laws:
((p q) (p q)) = T
= (pq)(¬pq) pq = ¬pq based on the propositional equivalence laws
= (p¬p)(qq) by expanding we get the following
= (p¬p)(¬qq) pq = ¬pq based on the propositional equivalence laws
= TT (¬pp) = T according to propositional laws
= T
Therefore
LHS = RHS so both are equivalent.
p
q
(p q)
(p q)
((p q) (p q))
0
0
1
0
1
0
1
0
0
0
1
0
0
0
0
1
1
1
1
1