最小化以下布尔表达式后的最小项数为______。
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[D′ + AB′ + A′C + AC′D + A′C′D]′
(A) 1. (B) 2. (C) 3. (D) 4. 答复: (A) 说明:
Given Boolean expression is: [D′ + AB′ + A′C + AC′D + A′C′D]′ Step 1 : [D′ + AB′ + A′C + C′D ( A + A')]′ ( taking C'D as common ) Step 2 : [D′ + AB′ + A′C + C′D]′ ( as, A + A' = 1 ) : [D' + DC' + AB' + A'C]' (Rearrange) Step 3 : [D' + C' + AB' + A'C]' ( Rule of Duality, A + A'B = A + B ) : [D' + C' + CA' + AB']' (Rearrange) Step 4 : [D' + C' + A' + AB']' (Rule of Duality) : [D' + C' + A' + AB']' (Rearrange) Step 5 : [D' + C' + A' + B']' (Rule of Duality) :[( D' + C' )'.( A' + B')'] (Demorgan's law, (A + B)'=(A'. B')) :[(D''.C'').( A''.B'')] (Demorgan's law) :[(D.C).(A.B)] (Idempotent law, A'' = A) : ABCD Hence only 1 minterm after minimization.
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