Mathematical Logic — Practice Quiz

∼q → ∼p, ∼p ∨ q, ∼(p ∧ ∼q)

Step 1: p → q is equivalent to ∼q → ∼p (contrapositive law). Step 2: p → q ≡ ∼p ∨ q (definition of implication). Step 3: Using De Morgan's law: ∼(p ∧ ∼q) ≡ ∼p ∨ q ≡ p → q. Step 4: q → p is the converse, not equivalent. Step 5: p ∧ q is conjunction, not equivalent to implication.

p ∨ ∼p, (p → q) ↔ (∼q → ∼p), p → (p ∨ q)

Step 1: p ∨ ∼p is always true (law of excluded middle) - tautology. Step 2: p ∧ ∼p is always false (contradiction). Step 3: (p → q) ↔ (∼q → ∼p) is contrapositive equivalence - tautology. Step 4: p → (p ∨ q): if p is true, then p ∨ q is true; if p is false, implication is true - tautology. Step 5: p ∧ q can be false when either p or q is false - not a tautology.

∼p ∨ ∼q

Step 1: De Morgan's law states ∼(p ∧ q) ≡ ∼p ∨ ∼q. Step 2: The negation of a conjunction becomes the disjunction of the negations. Step 3: This can be verified by truth table: when p ∧ q is false, at least one of p or q must be false, making ∼p ∨ ∼q true.

If the ground is not wet, then it does not rain

Step 1: Let p: 'it rains', q: 'the ground is wet'. Original: p → q. Step 2: Contrapositive of p → q is ∼q → ∼p. Step 3: ∼q: 'the ground is not wet', ∼p: 'it does not rain'. Step 4: Therefore: 'If the ground is not wet, then it does not rain'. Step 5: The contrapositive is logically equivalent to the original statement.