Unit 7 Review and Checkpoint
A free Logic lesson from the “Logical Equivalence” unit, with a worked example and practice problems including step-by-step solutions.
Equivalent statements agree on every assignment. Use truth tables or justified identities, not similar-looking wording. De Morgan swaps and/or and negates both parts. The implication p→q equals ¬p∨q. One disagreeing row disproves a claimed equivalence. This review checks performance on mathematical tasks. For extended written work, use the separate rubric with a tutor or teacher; the automatic quiz does not certify proof-writing proficiency.
What you'll learn
- Review equivalence, De Morgan's Laws, contrapositives, and truth-table tests
- Choose the reasoning tool that matches the statement
- Explain why an answer is valid, invalid, true, false, or unsupported
Worked example
Problem. Rewrite ¬(p ∧ q) using De Morgan's Law.
- Negate both parts and switch and to or.
Answer: ¬p ∨ ¬q
Lesson preview: this page shows 4 practice items. The interactive activity contains 4 practice items and 4 quiz items (multiple choice). Counts include repeated prompts and variants.
Practice problems
1. Rewrite ¬(p ∧ q) using De Morgan's Law.
Choices: ¬p ∨ ¬q · ¬p ∧ ¬q · p ∨ q · p ∧ q
Show solution
- Negate both parts and switch and to or.
Answer: ¬p ∨ ¬q
2. Rewrite p→q without an implication symbol.
Choices: p ∨ ¬q · p ∧ q · ¬p ∧ q · ¬p ∨ q
Show solution
- Both expressions are false exactly in the p=true, q=false row.
Answer: ¬p ∨ q
3. Simplify p ∧ (p ∨ q).
Choices: p ∧ q · ¬p · p · q
Show solution
- If p is false, the expression is false.
- If p is true, both factors are true regardless of q.
Answer: p
4. Which assignment shows p→q is not equivalent to q→p?
Choices: No such assignment exists · p=false, q=true · p=true, q=true · p=false, q=false
Show solution
- At this assignment p→q is true, while q→p is false.
Answer: p=false, q=true
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