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.

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

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