ON1 Big Mock Test 3 • October 2013 • Answers
This test counts 4% of assessment for the course.
Time allowed: 5 minutes
The actual test will contain only 2 questions.
Marking scheme: In each multiple choice question, 2 marks are given for a complete correct answer,
1 mark for an incomplete correct answer, 0 for an incorrect or partially incorrect answer or no answer.
A correct answer might contain more than one choice.
Tick the correct box (or boxes):
1. 1. Given that p is T , q is F and r is T , which of the following statements is F ?
(A) (p → (p → q)) ∨ r
(D) None of the above
(B) (p ∧ ∼ q) ↔ r
(C) (p ∧ ∼ q) →∼ r
Solution: (C). A direct computation of truth values. You need to remember truth
tables for conjunction ∧, disjunction ∨, conditional →, biconditional ↔.
2. Given that p is T , q is F and r is F , which of the following statements is T ?
(A) (p → q) ∨ r
(D) None of the above
(B) (p ∧ ∼ q) ↔ r
(C) (∼ p ∧ q) → r
Solution: (C). It is easy to compute truth values of each of the three formuale:
(A) (T → F ) ∨ F ≡ F ∨ F ≡ F ;
(B) (T ∧ ∼ F ) ↔ F ≡ (T ∧ T ) ↔ F ≡ T ↔ F ≡ F ;
(C) (∼ T ∧ F ) → F ≡ (F ∧ F ) → F ≡ F → F ≡ T .
3. Given that p → (q ∨ r) is F , which of the following statements is T ?
(A) ∼ q → (p ∧ r)
(D) None of the above
(B) (q → p) ∨ r
(C) (q ∧ r) ↔ p
Solution: From the truth table for →, the statement p → (q ∨ r) can be F only if p
is T and q → r is F .
But q ∨ r can be F only if q is F and r is F .
Hence we know the truth values of all simple statements:
p is T , q is F and r is F .
Now we can easily compute that the statements (A) and (C) are F and B is T . 4. Which of the following statements is a tautology (that is, takes logical value T for
every possible combination of logical values of p and q)?
(A) (p → q) ∨ (∼ p → q)
(B) (p ∧ q) ∨ (∼ p ∧ ∼ q)
(C) (q → p) ∨ (∼ p → ∼ q)
(D) All of the above
Solution: (A). Notice that the statement in (B) is F if p is T and q is F , while (C)
is false if q is T and p is F .
5. Which of the following statements is a tautology?
(A) p ∨ (p → q)
(C) (q → p) → (p → q)
(B) (p ∧ q) ∨ (∼ p ∧ ∼ q)
(D)None of the above
Solution: (A)
p ∨ (p → q) ≡
≡
≡
≡
p ∨ (∼ p ∨ q)
(p ∨ ∼ p) ∨ q
T ∨q
T.
6. Which of the following statements is a contradiction?
(A)
(q → p) → (p → q)
(C)
∼ (p ∧ ∼ p)
(B)
(p ∧ q) ∨ (∼ p ∧ ∼ q)
(D) None of the above
Solution: (D). All three formulae in (A)–(C) are T when p is T and q is T and
therefore are not contradictions.
7. Let X and Y be sets such of an universal set U . Which of the following must be
empty?
2
(A)
(C)
(X ∪ Y ) ∩ X 0 ∩ Y 0
(X ∩ U ) ∪ (Y ∩ U 0 )
(B) (X ∩ Y ) ∪ X 0 ∪ Y 0
(D) None of the above
Solution: (A). This can be seen from a simple Boolean algebra calculation:
(X ∪ Y ) ∩ X 0 ∩ Y 0 = (X ∪ Y ) ∩ (X 0 ∩ Y 0 ) (associativity of ∩)
= (X ∪ Y ) ∩ (X ∪ Y )0 (De Morgan’s Law)
= ∅,
or from drawing a Venn diagram.
The set in (B) is non-empty when, for example, X = ∅ and U is non-empty:
(X ∩ Y ) ∪ X 0 ∪ Y 0 = (∅ ∩ Y ) ∪ ∅0 ∪ Y 0 = ∅ ∪ U ∪ Y 0 = U.
The set in (B) is non-empty when, for example, X = U and U is non-empty:
(X ∩ U ) ∪ (Y ∩ U 0 ) = (U ∩ U ) ∪ (Y ∩ ∅) = U ∪ ∅ = U.
8. Which of the following statements is logically equivalent to ∼ p ∨ (p ∧ q)?
(A)
q→p
(B)
∼p∨q
(C)
∼ (p ∨ q)
(D) None of he above
Solution: the problem can be solved in three different ways:
1. By constructing truth tables for all formulae – it is not that difficult. Indeed,
there is no need to fill in the whole table; as soon as you see that the two tables are
different, the formulae are not logically equivalent.
2. Alternatively, one may use basic logical equivalences:
∼ p ∨ (p ∧ q) ≡ (∼ p ∨ p) ∧ (∼ p ∨ q) ≡ T ∧ (∼ p ∨ q) ≡∼ p ∨ q.
Therefore the correct answer is (B).
3. One may directly observe that for ∼ p ∨ (p ∧ q) to be T , either p has to be F ,
or both p and q has to be true (alternatively, construct the truth table!). Therefore
∼ p ∨ (p ∧ q) has the same truth values as p → q. But we know that
p → q ≡∼ p ∨ q.
Therefore the correct answer is (B).
9. Which of the following statements is logically equivalent to (p ∧ q)∨ ∼ p?
(A) p → q
(B) p ∨ ∼ q
(C) ∼ (p ∧ q)
3
(D) None of he above
Solution: The correct answer is A.
(p ∧ q)∨ ∼ p ≡
≡
≡
≡
≡
(p∨ ∼ p) ∧ (q∨ ∼ p)
(distributivity)
T ∧ (q∨ ∼ p)
(truth table for ∨)
(q∨ ∼ p)
(truth table for ∧)
(∼ p ∨ q)
(commutativity of ∨)
(p → q)
(from the lectures)
Alternatively (and perhaps simpler): just compute truth tables for every formula.
10. Which of the following statements is logically equivalent to
p ∧ ∼ (p → q)?
(A) q → p
(B) p ∧ ∼
(C) F
q
(D) None of he above
Solution: (B). There are two possible solutions—one is to compute truth tables for
all formulae in the problem, the other is to use fundamental logical equivalences. I
give here the second solution:
p ∧ ∼ (p → q)
by (8)
≡
by (7)
≡
by (6)
≡
by (3)
≡
by (2)
≡
p∧ ∼ (∼ p ∨ q)
p ∧ (∼∼ p ∧ ∼ q)
p ∧ (p ∧ ∼ q)
(p ∧ p) ∧ ∼ q
p ∧ ∼ q.
4