Solutions to the Homework I
I601 Logic and Discrete Mathematics
1.
Determine whether each of these compound propositions is satisfiable.
a) (p ∨ ¬q) ∧ (¬p ∨ q) ∧ (¬p ∨ ¬q)
b) (p → q) ∧ (p → ¬q) ∧ (¬p → q) ∧ (¬p → ¬q)
c) (p ↔ q) ∧ (¬p ↔ q)
S o l u t i o n. For part (a) we have the following table
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
p ∨ ¬q
T
T
F
T
¬p ∨ q
T
F
T
T
¬p ∨ ¬q
F
T
T
T
(p ∨ ¬q) ∧ (¬p ∨ q) ∧ (¬p ∨ ¬q)
F
F
F
T
From the last column we see that the proposition is satisfiable.
The same result can be obtained using logically equivalent transformations:
(p ∨ ¬q) ∧ (¬p ∨ q) ∧ (¬p ∨ ¬q) ≡ ((p ∧ (¬p ∨ q)) ∨ (¬q ∧ (¬p ∨ q))) ∧ (¬p ∨ ¬q)
≡ ((p ∧ ¬p) ∨ (p ∧ q)) ∨ (¬q ∧ ¬p) ∨ (¬q ∧ q)) ∧ (¬p ∨ ¬q)
≡ ((p ∧ q)) ∨ (¬p ∧ ¬q)) ∧ (¬p ∨ ¬q)
≡ ¬p ∧ ¬q
This shows that the proposition is true if both p and q are false. Thus, the proposition is satisfiable.
Analogously, we can use method of truth tables for parts (b) and (c).
Part (b): Considering that
(p → q) ∧ (p → ¬q) ∧ (¬p → q) ∧ (¬p → ¬q) ≡ (¬p ∨ q) ∧ (¬p ∨ ¬q) ∧ (¬¬p ∨ q) ∧ (¬¬p ∨ ¬q)
≡ (¬p ∨ q) ∧ (¬p ∨ ¬q) ∧ (p ∨ q) ∧ (p ∨ ¬q)
we can compose the truth table as follows:
p
T
T
F
F
q
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
¬p ∨ q
T
F
T
T
¬p ∨ ¬q
F
T
T
T
p∨q
T
T
T
F
p ∨ ¬q
T
T
F
T
Hence, this proposition is not satisfiable (it is the contradiction).
Entire proposition
F
F
F
F
Part (c): The truth table as follows:
p q
T T
T F
F T
F F
Step:
(p ↔ q)
T
F
F
T
1
∧
F
F
F
F
3
(¬p ↔ q)
F
T
T
F
2
Hence, this proposition is not satisfiable.
2. Use the laws in the lecture slide #28 to simplify the formula ¬(p ∧ q) ∨ (¬p ∧ q).
Check the results using the technique of truth tables.
S o l u t i o n.
¬(p ∧ q) ∨ (¬p ∧ q) ≡ ¬p ∨ ¬q ∨ (¬p ∧ q)
≡ ¬p ∨ ((¬q ∨ ¬p) ∧ (¬q ∨ q))
≡ ¬p ∨ ((¬q ∨ ¬p) ∧ T)
≡ ¬p ∨ (¬q ∨ ¬p)
≡ ¬q ∨ ¬p
≡ p → ¬q
De Morgan law
associative and distributive laws
complement law
identity law
assiociative and idempotent laws
Example 2 on slide #24
Alternative answers may be ¬q ∨ ¬p or q → ¬p.
To check the result one can compose the following truth table:
p q
T T
T F
F T
F F
Step:
¬
F
T
T
T
2
(p ∧ q)
T
F
F
F
1
∨
F
T
T
T
5
(¬p
F
F
T
T
3
∧q)
F
F
T
F
4
≡
T
T
T
T
8
p→
F
T
T
T
7
¬q
F
T
F
T
6
3. Find CNF (Conjunctive Normal Form) and DNF (Disjunctive Normal Form)
for the formula
(p → q) ↔ r ∧ ¬p.
S o l u t i o n.
(p → q) ↔ r ∧ ¬p ≡ [(p → q) → r ∧ ¬p] ∧ [r ∧ ¬p → (p → q)]
≡ [(¬p ∨ q) → r ∧ ¬p] ∧ [r ∧ ¬p → (¬p ∨ q)]
≡ [¬(¬p ∨ q) ∨ (r ∧ ¬p)] ∧ [¬(r ∧ ¬p) ∨ ¬p ∨ q]
≡ [(p ∧ ¬q) ∨ (r ∧ ¬p)] ∧ [¬(r ∧ ¬p) ∨ ¬p ∨ q]
≡ [(p ∧ ¬q) ∨ (r ∧ ¬p)] ∧ (¬r ∨ p ∨ ¬p ∨ q)
≡ [(p ∧ ¬q) ∨ (r ∧ ¬p)] ∧ (¬r ∨ T ∨ q)
≡ [(p ∧ ¬q) ∨ (r ∧ ¬p)] ∧ T
≡ (p ∧ ¬q) ∨ (r ∧ ¬p)
≡ (p ∨ r) ∧ (¬q ∨ r) ∧ (¬q ∨ ¬p)
def. of biconditional
Example 2 on slide #24
Example 2 on slide #24
De Morgan law
De Morgan law
complement law
domination law
domination law
distributive law
The last two lines yield DNF and CNF respectively:
DNF: (p ∧ ¬q) ∨ (r ∧ ¬p)
CNF: (p ∨ r) ∧ (¬q ∨ r) ∧ (¬p ∨ ¬q)
4. A collection of logical operators is called functionally complete if every compound proposition is logically equivalent to a compound proposition involving only
these logical operators. For example, as every compound proposition is convertible
to CNF , the connectives ¬, ∧ and ∨ form a functionally complete collection of logical
operators.
Show that ¬ and ∧ form a functionally complete collection of logical operators.
Find a formula containing propositional variables p, q and r, that is true if and only
if exactly one of those variables is true. Represent this formula in the form using
only connectives ¬ and ∧.
S o l u t i o n. Every proposition can be converted into DNF or CNF containing connectives
∨, ∧ and ¬. Using the De Morgan law, one can substitute every ∨ by its logical equivalent:
p ∨ q ≡ ¬(¬p ∧ ¬q).
For example, the proposition that is true iff exactly one of those variables is true has three
alternatives when it is valid:
(a) only p is true: p ∧ ¬q ∧ ¬r ≡ T
(b) only q is true: ¬p ∧ q ∧ ¬ ≡ T
(c) only r is true: ¬p ∧ ¬q ∧ r ≡ T
So, the proposition can be written as
(p ∧ ¬q ∧ ¬r) ∨ (¬p ∧ q ∧ ¬r) ∨ (¬p ∧ ¬q ∧ r).
Using distributive and absorption laws several times, we see that this formula is logically
equivalent to
(p ∨ q ∨ r) ∧ (¬q ∨ ¬r) ∧ (¬p ∨ ¬r) ∧ (¬p ∨ ¬q)
(1)
This can be converted using the De Morgan law into the form containing only connectives ¬ and
∧:
¬(¬p ∧ ¬q ∧ ¬r) ∧ ¬(q ∧ r) ∧ ¬(p ∧ r) ∧ ¬(p ∧ q).
To verify the formula against the initial requirements, we construct the truth table of the
proposition (1):
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 ∨ r)
T
T
T
T
T
T
T
F
1
∧
F
T
T
T
F
T
T
F
5
(¬q ∨ ¬r)
F
T
T
T
F
T
T
T
2
∧
F
T
F
T
F
T
T
F
6
(¬p ∨ ¬r)
F
T
F
T
T
T
T
T
3
∧
F
F
F
T
F
T
T
F
7
(¬p ∨ ¬q)
F
F
T
T
T
T
T
T
4