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De Morgan's Laws:

~(A /\ B) <=> ~A \/ ~B

~(A \/ B) <=> ~A /\ ~B

Given the statement, rewrite using De Morgan's Laws, noting that ~(~P) should be equivalently written as P.

~(~A \/ B) is equivalent to?

Give the answer using capital letters and no spaces, no parentheses.

Copy and Paste the symbols as needed:

∨ ∧ ⇒

De Morgan's Laws: ~(A /\ B) <=> ~A \/ ~B ~(A \/ B) <=> ~A /\ ~B Given-example-1
User Jeff Burka
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1 Answer

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Final answer:

Using De Morgan's Laws, the expression ~(~A ∨ B) is altered by negating both terms independently and switching the disjunction to a conjunction, which simplifies to AB when unnecessary negations are removed.

Step-by-step explanation:

The expression ~(~A ∨ B) can be rewritten using De Morgan's Laws. According to De Morgan's Laws, the negation of a disjunction (~A ∨ ~B) is equivalent to the conjunction of the negations (~A ∧ ~B), and conversely, the negation of a conjunction (A ∧ B) is equivalent to the disjunction of the negations (~A ∨ ~B). Therefore, applying this rule to the given statement:

~(~A ∨ B) is equivalent to applying the negation to both terms independently and changing the disjunction to a conjunction:

¬(~A) ∧ ¬B

Recall that two negatives cancel out, so ¬(~A) is equivalent to A. Thus, the rewritten expression using De Morgan's Laws without spaces or parentheses is:

AB

User OpenStack
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