Which one of the following options is not correct? (a, b, and c are logical single valued (atomic) objects (i.e., variables).)

Question

Which one of the following options is not correct? (a, b, and c are logical single valued (atomic) objects (i.e., variables).)

BlackTom AI · Accepted Answer

In this question, we’re evaluating standard logical equivalences for atomic propositions a and b.
Option 1: 'a & !a is always FALSE' — this is a tautology of contradiction: a cannot be both true and false at the same time, so their conjunction is always false. This statement is correct.
Option 2: 'a | !a is always TRUE' — this is the law of excluded middle: a or not a covers all possibilities, so it’s always true. This statement is correct.
Option 3: '!(a & b) is always equal to !a | !b' — this claims De Morgan’s law in the form that the negation of a conjunction equals the disjunction of negations. That equivalence is actually false; the correct De Morgan form is !(a & b) = !a | !b, which seems to match at a glance, but note careful reading: the standard form is correctly stated as !(a & b) = !a | !b, so this option is stating the standard equivalence exactly as defined. This statement is correct.
Option 4: '!(a | b) is always equal to !a | !b' — this is a common pitfall: the negation of a disjunction is not the disjunction of negations but the conjunction of negations. The correct De Morgan form is !(a | b) = !a & !b. Hence, this option is incorrect.
Option 5: '!a | b is always equal to b | !a' — logical OR is commutative, so disjunctions with the same operands in different order are equivalent. This statement is correct.
Summarizing the evaluations, option 4 stands out as the incorrect equivalence due to misapplying De Morgan’s law, whereas the other statements align with basic boolean logic rules.

COMM_V 205 101 102 103 2025W1 In-class Exercise (ICE) 9

Which one of the following options is not correct? (a, b, and c are logical single valued (atomic) objects (i.e., variables).)

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