Unit 2 Cycle 2 Day 2: De Morgan's with Multiple Operators
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De Morgan's with Multiple Operators
Section 2.6 — Equivalent Boolean
Key Concept
When De Morgan's Laws are applied to expressions with more than two operands, apply the transformation systematically. For !(a && b || c), first respect operator precedence: this is !((a && b) || c), which becomes !(a && b) && !c, then (!a || !b) && !c. The AP exam adds complexity with mixed relational operators: negating x > 5 && x < 10 gives x <= 5 || x >= 10. Each relational operator flips to its complement when negated.
Consider the following boolean expression.
Which of the following is equivalent to the expression above?
Answer: (A) a <= 5 && (b > 3 || c == 0)
Apply De Morgan's step by step: !(a > 5 || (b <= 3 && c != 0)) becomes !(a > 5) && !(b <= 3 && c != 0) (negate both, flip OR to AND). Then a <= 5 && (!(b <= 3) || !(c != 0)) = a <= 5 && (b > 3 || c == 0).
Why Not the Others?
(B) The outer OR should be AND (De Morgan's flips OR to AND when negated).
(C) This incorrectly distributes the negation without applying De Morgan's to the inner expression.
(D) The inner OR (from negating the inner AND) is missing. b > 3 and c == 0 should be connected by OR, not AND.
Common Mistake
Apply De Morgan's from outside in. First negate the outer OR to get AND. Then negate the inner AND to get OR. Negate each leaf expression by flipping the relational operator.
AP Exam Tip
For complex De Morgan's, work in layers. Negate the outermost operator first, then recurse into sub-expressions. Keep parentheses to maintain grouping.