LOGIC QUESTIONS WITH SOLUTIONS
Tautology: A compound statement that is always true, regardless of the truth values of the individual statements.
Contradiction: A compound statement that is always false, regardless of the truth values of the individual statements.
Example:
p ∨ ¬p is a tautology (Law of Excluded Middle)
p ∧ ¬p is a contradiction (Law of Contradiction)
(ii) Show that (p ∧ q) ∧ (¬p ∨ q) is a fallacy using truth table
| p | q | ¬p | p ∧ q | ¬p ∨ q | (p ∧ q) ∧ (¬p ∨ q) |
|---|---|---|---|---|---|
| T | T | F | T | T | T |
| T | F | F | F | F | F |
| F | T | T | F | T | F |
| F | F | T | F | T | F |
The statement is only true in one case out of four (when both p and q are true), therefore it is not a tautology and constitutes a fallacy.
"On my daughter's birthday, I bring her flowers. Either it is my daughter's birthday or I work late. I did not bring her flowers. Therefore I worked late."
(i) Translate to symbolic form
Let:
B = "It is my daughter's birthday"
F = "I bring her flowers"
W = "I work late"
Premises:
1. B → F
2. B ∨ W
3. ¬F
Conclusion: W
(ii) Test validity using truth table
| B | F | W | B→F | B∨W | ¬F | Valid? |
|---|---|---|---|---|---|---|
| T | T | T | T | T | F | - |
| T | T | F | T | T | F | - |
| T | F | T | F | T | T | ✗ (Premise 1 false) |
| T | F | F | F | T | T | ✗ (Premise 1 false) |
| F | T | T | T | T | F | - |
| F | T | F | T | F | F | ✗ (Premise 2 false) |
| F | F | T | T | T | T | ✓ (All premises true, conclusion true) |
| F | F | F | T | F | T | ✗ (Premise 2 false) |
The only case where all premises are true is when B is false, F is false, and W is true. In this case, the conclusion W is true, so the argument is valid.
| p | q | p ∧ q | p ∨ q | (p ∧ q) → (p ∨ q) |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | F | T | T |
| F | T | F | T | T |
| F | F | F | F | T |
The implication is true in all possible cases, therefore it is a tautology.
(i) (P ∨ q) ∧ [¬q ∨ ¬r) ∧ (P ∨ r)]
Solution:
Let's simplify step by step:
1. Apply Distributive Law: (A ∧ B) ∨ (A ∧ C) = A ∧ (B ∨ C)
2. The expression is already in a distributed form. We can analyze cases:
- If P is true, the whole expression simplifies to ¬q ∨ ¬r
- If q is true, the ¬q term becomes false
- If r is true, the ¬r term becomes false
Final simplified form: P ∧ (¬q ∨ ¬r)
(ii) (P ∨ q) ∧ ¬(¬P ∧ q)
Simplify using De Morgan's laws:
¬(¬P ∧ q) = P ∨ ¬q (De Morgan's)
Now we have: (P ∨ q) ∧ (P ∨ ¬q)
Apply Distributive Law: P ∨ (q ∧ ¬q) = P ∨ F = P
Original statement: "If Juma is a poet, then he is poor"
Let:
P = "Juma is a poet"
Q = "Juma is poor"
Original: P → Q
Converse: Q → P ("If Juma is poor, then he is a poet")
Contrapositive: ¬Q → ¬P ("If Juma is not poor, then he is not a poet")
Inverse: ¬P → ¬Q ("If Juma is not a poet, then he is not poor")
Note: The contrapositive is logically equivalent to the original statement.
Key Logic Concepts Covered:
- Tautologies and contradictions
- Truth table analysis
- Argument validity testing
- Propositional simplification using laws of logic
- Converse, contrapositive and inverse statements
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