Solutions to the Quiz 1 (2016)
I601 Logic and Discrete Mathematics
A1. Let p, q, r denote primitive propositions. Determine whether the statements
(p ∨ q) → r and (p → r) ∧ (q → r) are logically equivalent.
S o l u t i o n. The truth table for the formula [(p ∨ q) → r] ↔ [(p → r) ∧ (q → r)]:
p
T
T
T
T
F
F
F
F
q r
T T
T F
F T
F F
T T
T F
F T
F F
Step:
[(p ∨ q)
T
T
T
T
T
T
F
F
1
→ r]
T
F
T
F
T
F
T
T
2
↔
T
T
T
T
T
T
T
T
6
[(p → r)
T
F
T
F
T
T
T
T
3
∧
T
F
T
F
T
F
T
T
5
(q → r)]
T
F
T
T
T
F
T
T
4
The column corresponding to the sixth step shows that the formulae represent logically
equivalent propositions.
A l t e r n a t i v e s o l u t i o n. Using the laws of logic we obtain:
(p ∨ q) → r ≡ ¬(p ∨ q) ∨ r
≡ (¬p ∧ ¬q) ∨ r
≡ (¬p ∨ r) ∧ (¬q ∨ r)
≡ (p → r) ∧ (q → r)
implication
De Morgan law
distributive law
implication
B1. Let p, q, r denote primitive propositions. Determine whether the statements
p → q ∧ r and (p → q) ∧ (p → r) are logically equivalent.
S o l u t i o n. The truth table for the formula [p → q ∧ r] ↔ [(p → q) ∧ (p → r)]:
p
T
T
T
T
F
F
F
F
q r
T T
T F
F T
F F
T T
T F
F T
F F
Step:
p→
T
F
F
F
T
T
T
T
1
q∧r
T
F
F
F
T
F
F
F
2
↔
T
T
T
T
T
T
T
T
6
[(p → q)
T
T
F
F
T
T
T
T
3
∧
T
F
F
F
T
T
T
T
5
((p → r)]
T
F
T
F
T
T
T
T
4
The column corresponding to the sixth step shows that the formulae represent logically
equivalent propositions.
A l t e r n a t i v e s o l u t i o n. Using the laws of logic we obtain:
p→q∧r
≡ ¬p ∨ (q ∧ r)
≡ (¬p ∨ q) ∧ (¬p ∨ r)
≡ (p → q) ∧ (p → r)
implication
distributive law
implication
A2. Construct the truth table for the statement (p → q) ∧ (p → ¬q). Propose a simpler statement (containing no more than one connective) that is logically equivalent
to this formula.
S o l u t i o n. This formula has the following truth table
p q
T T
T F
F T
F F
Step:
(p → q)
T
F
T
T
2
∧
F
F
T
T
4
(p →
F
T
T
T
3
¬q)
F
T
F
T
1
The column corresponding to the forth step shows that the values of the proposition are just
inverse to values of p in the first column. Hence, the formula is logically equivalent to the
proposition ¬p.
To verify this result, one may transform the statement as follows:
(p → q) ∧ (p → ¬q) ≡ (¬p ∨ q) ∧ (¬p ∨ ¬q)
≡ ¬p ∨ (q ∧ ¬q)
≡ ¬p ∨ T
≡ ¬p
implication
distributive law
complement law
identity law
B2. Construct the truth table for the statement p → ((¬p → q) → ¬q). Propose
a simpler statement (containing no more than two connectives) that is logically
equivalent to this formula.
S o l u t i o n. This formula has the following truth table
p q
T T
T F
F T
F F
Step:
p→
F
T
T
T
5
((¬p
F
F
T
T
1
→ q)
T
T
T
F
2
→
F
T
F
T
4
¬q)
F
T
F
T
3
The column corresponding to the fifth step is inverse to the conjunction of p and q. Hence, the
proposition is logically equivalent to ¬(p ∧ q). The latter is logically equivalent to ¬p ∨ ¬q or
p → ¬q or q → ¬p.
To verify this result, one may transform the statement as follows:
p → ((¬p → q) → ¬q) ≡ p → ((p ∨ q) → ¬q)
≡ p → (¬(p ∨ q) ∨ ¬q)
≡ p → ((¬p ∧ ¬q) ∨ ¬q)
≡ p → ¬q
implication
implication
De Morgan law
absorption law