The number of min-terms after minimizing the following Boolean expression isβ¦
2015
The number of min-terms after minimizing the following Boolean expression is _________.
[π·β² + π΄π΅β² + π΄β²πΆ + π΄πΆβ²π· + π΄β²πΆβ²π·]β²
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Correct answer: 1
Key idea: apply De Morgan's law to the complement and simplify algebraically.
Apply De Morgan: F = [D' + AB' + A' C + A C' D + A' C' D]' = D * (AB')' * (A' C)' * (A C' D)' * (A' C' D)'.
Take complements of products: (AB')' = A' + B, (A' C)' = A + C', (A C' D)' = A' + C + D', (A' C' D)' = A + C + D'.
Because F has an overall factor D, use D*(X + D') = D*X to remove the +D' terms: F = D*(A' + B)*(A + C')*(A' + C)*(A + C).
Use (A + C)(A + C') = A, so F = D * A * (A' + B) * (A' + C).
Simplify A*(A' + B) = AB, then AB*(A' + C) = ABC, giving F = D * ABC = ABCD.
Final result: the minimized expression is ABCD, which is a single minterm. Therefore the number of minterms is 1.
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