AP CSA Unit 2.6: De Morgan's Law and Equivalent Expressions Practice
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Unit 2, Section 2.6
Day 6 Practice • January 12, 2026
🎯 Focus: De Morgan's Law
Practice Question
Consider the following two boolean expressions:
// Expression 1
!(a && b)
// Expression 2
!a || !bFor which values of
a and b do Expression 1 and Expression 2 produce different results?
What This Tests: De Morgan's Law states that !(a && b) is equivalent to !a || !b. These expressions ALWAYS produce the same result for any values of a and b.
De Morgan's Laws
// De Morgan's Laws - MEMORIZE THESE!
!(a && b) ≡ !a || !b
!(a || b) ≡ !a && !b
// Memory trick: "Distribute NOT, flip the operator"
// NOT-AND becomes OR-NOT
// NOT-OR becomes AND-NOTTruth Table Proof
| a | b | !(a && b) | !a || !b | Same? |
|---|---|---|---|---|
| T | T | F | F | ✓ |
| T | F | T | T | ✓ |
| F | T | T | T | ✓ |
| F | F | T | T | ✓ |
All rows match! The expressions are logically equivalent.
Practice Technique
When to Use De Morgan's
De Morgan's Law helps simplify complex conditions:
if (!(x > 5 && y < 10))
becomes:
if (x <= 5 || y >= 10)
Difficulty: Medium • Time: 2-3 minutes • AP Skill: 2.B - Equivalent expressions
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