最小化以下布尔表达式后的最小项数为_________。
[D′ + AB′ + A′C + AC′D + A′C′D]′ (A) 1
(B) 2
(C) 3
(D) 4答案： (A)
解释：
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. 这个问题的测验