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Name: ____________________ Date: ____________
1.Easy
Exactly one row of the truth table makes (b ∧ ¬a) ∧ (c ∧ (¬b → c)) true. Click that row.
(b ∧ ¬a) ∧ (c ∧ (¬b → c))
1a = 1, b = 1, c = 1
2a = 1, b = 1, c = 0
3a = 1, b = 0, c = 1
4a = 1, b = 0, c = 0
5a = 0, b = 1, c = 1
6a = 0, b = 1, c = 0
7a = 0, b = 0, c = 1
8a = 0, b = 0, c = 0
2.Easy
Exactly one row of the truth table makes p → (p ∧ q) false. Click that row.
p → (p ∧ q)
1p = 1, q = 1
2p = 1, q = 0
3p = 0, q = 1
4p = 0, q = 0
3.Medium
You want to prove q → (r ∨ p) by contradiction. Which statement do you assume at the start? Click it.
q → (r ∨ p)
1¬q → ¬(r ∨ p)
2¬q ∧ ¬(r ∨ p)
3¬q ∧ (r ∨ p)
4q ∧ ¬(r ∨ p)
4.Medium
Is a ∧ ¬a a tautology, a contradiction or a contingency? Click your answer.
a ∧ ¬a
1tautology
2contradiction
3contingency
5.Hard
How many of the 16 rows of the truth table make ((d → c) ∧ (b ∨ ¬c)) → ((b ∨ ¬d) ∧ a) true? Call this number N.
((d → c) ∧ (b ∨ ¬c)) → ((b ∨ ¬d) ∧ a)
N =
6.Hard
The argument with premises q → (r ∨ p); q and conclusion r is invalid. Click the row of the truth table that is a counterexample.
q → (r ∨ p)q∴ r
1p = 1, q = 1, r = 1
2p = 1, q = 1, r = 0
3p = 1, q = 0, r = 1
4p = 1, q = 0, r = 0
5p = 0, q = 1, r = 1
6p = 0, q = 1, r = 0
7p = 0, q = 0, r = 1
8p = 0, q = 0, r = 0
Answer key
1. a = 0, b = 1, c = 1
2. p = 1, q = 0
3. q ∧ ¬(r ∨ p)
4. contradiction
5. N = 12
6. p = 1, q = 1, r = 0
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