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Name: ____________________ Date: ____________
1.Easy
Exactly one row of the truth table makes (¬b ↔ (b ∧ ¬a)) ∧ (a ↔ c) true. Click that row.
(¬b ↔ (b ∧ ¬a)) ∧ (a ↔ 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 ¬((b ∨ a) → (b ↔ ¬a)) true. Click that row.
¬((b ∨ a) → (b ↔ ¬a))
1a = 1, b = 1
2a = 1, b = 0
3a = 0, b = 1
4a = 0, b = 0
3.Medium
Which formula is logically equivalent to ¬(c ↔ a)? Click it.
¬(c ↔ a)
1c ∧ ¬a
2¬c ↔ ¬a
3c ↔ ¬a
4¬c ∨ ¬a
4.Medium
In how many of the 8 rows of its truth table is (q ∧ (p → r)) ∧ (p → ¬(q ∧ r)) true? Call this number N.
(q ∧ (p → r)) ∧ (p → ¬(q ∧ r))
N =
5.Hard
The argument with premises b → a; c → a and conclusion b → c is invalid. Click the row of the truth table that is a counterexample.
b → ac → a∴ 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
6.Hard
Which statement is equivalent to ¬∃x (F(x) ∧ G(x)), with the negation moved all the way inside? Click it.
¬∃x (F(x) ∧ G(x))
1∀x (¬F(x) ∧ ¬G(x))
2∃x (¬F(x) ∧ ¬G(x))
3∃x (F(x) → ¬G(x))
4∀x (F(x) → ¬G(x))
Answer key
1. a = 1, b = 1, c = 1
2. a = 1, b = 1
3. c ↔ ¬a
4. N = 2
5. a = 1, b = 1, c = 0
6. ∀x (F(x) → ¬G(x))
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