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