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The number of min-terms after minimizing the following Boolean expression is _______.

$[D'+AB'+A'C+AC'D+A'C'D]'$
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$F=\overline{[\overline{D}+A\overline{B}+\overline{A}C+A\overline{C}D+\overline{A}\ \overline{C}D]}$

This is already arranged in a beautifully symmetric complemented SOP form, let's not mess up this symmetry by minimising. Instead, we can do this:

$\overline{F}=[\overline{D}+A\overline{B}+\overline{A}C+A\overline{C}D+\overline{A}\ \overline{C}D]$

 

We know that minterms have the functional output 1.

Here, these minterms have the functional output 1 for $\overline{F}$

Hence, these same minterms have functional output 0 for $F$

 

So, K-map for $F$

Minterms = 1


Alternate method

You can draw K-map for $\overline{F}$ and then to fetch K-map of $F$, interconvert all the $1's$ and $0's$

K-map for $\overline{F}$

K-map for $\overline{F}$ has one 0. So K-map for $F$ will have one 1.

So, 1

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[D'+AB'+A'C+AC'D+A'C'D]'

After applying complementation on given expression we get

>>D(A'+B)(A+C')(A'+C+D')(A+C+D')

>>(A'D+BD)(A+C')(A'+C+D')(A+C+D')

>>(A'AD+A'C'D+ABD+BC'D)(A'A+A'C+A'D'+AC+C+CD'+AD'+CD'+D')

>>(0+A'C'D+ABD+BC'D)(0+A'C+A'D'+AC+C+CD'+AD'+CD'+D')

>>(A'C'D+ABD+BC'D)(A'C+AC+CD'+C+AD'+A'D'+D')   

>>(A'C'D+ABD+BC'D)(C+D')  

>>ABCD only minterm

 

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$ [D′ + AB′ + A′C + AC′D + A′C′D]′$

 making it canonical SOP

D′ it has missing ABC and they can be in 8 possible combinations with D' = 0 at end

xxxD’

xxx0

0000  0

0010  2

0100  4

0110  6

1000  8

1010  10

1100  12

1110  14

AB′ has missing CD and it will have four combination with 

AB′xx

10xx

1000  8

1001  9

1010  10

1011  11

A′C has missing BD  and it will have four combination with 

A′xCx

0x1x

0010  2

0011  3

0110  6

0111  7

AC′D has missing B and it will have twocombination with

AxC′D

1x01

0001  1

0101  5

A′C′D has missing B and it will have twocombination with

A′xC′D

0x01

0001  1

0101  5

$=Σ(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14)’$ 

$= Σ(15) =ABCD $

i.e. only 1 term 

 

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[D′+AB′+A′C+AC′D+A′C′D]′
=> [AC’D+A′C′D+D′+AB’+A’C]’
=>[AC’+A’C’+D’+AB’+A’C]’              {applying postulate X+X’Y== X’+Y twice}
=>[C’+A’(C+C’)+ D’ +AB’]’
=>[C’+A’+AB’+D’]’
=>[A’+B’+C’+D’]’
=>(A’)’ (B’)’(C’)’(D’)’                     {since (x+y)’= x’y’}
=>ABCD

Answer:

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