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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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Best answer
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75 votes

$F = [D'+AB'+A'C+AC'D+A'C'D]'$

$F'= D'+AB'+A'C+AC'D+A'C'D$

Now we have F', so fill 0's (maxterms) in K-map for each term

As for D'

Similarly for $AB'$, $A'C, AC'D$ and $A'C'D$.  We will get

We get one place for minterm and that is ABCD

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Let's First Simplify it 

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

[D' + AB' + A'C + C'D (A + A') ]'     // A + A' =1

[AB' +A'C + (D' + C') (D' + D) ]'    // Apply Distributive Rule Among D' and C'D

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

[AB' + (A' + C') (C + C') + D'  ] '   //Apply distributive Law B/w A'C and C'

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

[(A+ A') (A' + B') +  C' + D' ]'    //Apply Distributive law b/w AB' and A'

Finally we got

[A' +B' +C' + D' ]'

Apply Demorgan's Law

ABCD

19 votes
19 votes
Canonical SOP of the expression D'+AB'+A'C+AC'D+A'C'D is ∑m(0,1,2,3,4,5,6,7,9,13,12,14,11,8,10)

So corresponding POS form is ⫪M(15)

minterm=(maxterm)'

minterm=(A'+B'+C'+D')'=ABCD

Therefore number of minterms = 1
19 votes
19 votes

$F=[D'+AB'+A'C+AC'D+A'C'D]'$

$F'=D'+AB'+A'C+AC'D+A'C'D$

$F'=1.D'+AB'.1+A'.1C+A.1C'D+A'.1C'D$

$F'=(A+A')D'+AB'(C+C')+A'(B+B')C+A(B+B')C'D+A'(B+B')C'D$

$F'=AD'+A'D'+AB'C+AB'C'+A'BC+A'B'C+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=A.1D'+A.1'D'+AB'C.1+AB'C'.1+A'BC.1+A'B'C.1+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=A(B+B')D'+A'(B+B')D'+AB'C(D+D')+AB'C'(D+D')+A'BC(D+D')+A'B'C(D+D')+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=ABD'+AB'D'+A'BD'+A'B'D'+AB'CD+AB'CD'+AB'C'D+AB'C'D'+A'BCD+A'BCD'+A'B'CD+A'B'CD'+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=AB.1D'+AB'.1D'+A'B.1D'+A'B'.1D'+AB'CD+AB'CD'+AB'C'D+AB'C'D'+A'BCD+A'BCD'+A'B'CD+A'B'CD'+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=AB(C+C')D'+AB'(C+C')D'+A'B(C+C')D'+A'B'(C+C')D'+AB'CD+AB'CD'+AB'C'D+AB'C'D'+A'BCD+A'BCD'+A'B'CD+A'B'CD'+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

$F'=ABCD'+ABC'D'+AB'CD'+AB'C'D'+A'BCD'+A'BC'D'+A'B'CD'+A'B'C'D'+AB'CD+AB'CD'+AB'C'D+AB'C'D'+A'BCD+A'BCD'+A'B'CD+A'B'CD'+ABC'D+AB'C'D+A'BC'D+A'B'C'D$

Remove similar tearm,because $[X+X=X]$

$F'=ABCD'+ABC'D'+AB'CD'+AB'C'D'+A'BCD'+A'BC'D'+A'B'CD'+A'B'C'D'+AB'CD+AB'C'D+A'BCD+A'B'CD+ABC'D+A'BC'D+A'B'C'D$

Apply both side complement

$(F')'=[ABCD'+ABC'D'+AB'CD'+AB'C'D'+A'BCD'+A'BC'D'+A'B'CD'+A'B'C'D'+AB'CD+AB'C'D+A'BCD+A'B'CD+ABC'D+A'BC'D+A'B'C'D]'$

Apply Demorgan's laws

$(A+B)'=A'.B'$

$(A.B)'=A'+B'$

$F=(A'+B'+C'+D).(A'+B'+C+D).(A'+B+C'+D).(A'+B+C+D).(A+B'+C'+D).(A+B'+C+D).(A+B+C'+D).(A+B+C+D).(A'+B+C'+D').(A'+B+C+D').(A+B'+C'+D').(A+B+C'+D').(A'+B'+C+D').(A+B'+C+D').(A+B+C+D')$                 $[(X')'=X]$

This is Canonical Product Of Sum Term(Maxterm)

$1)A'+B'+C'+D=1110=14$

$2)A'+B'+C+D=1100=12$

$3)A'+B+C'+D=1010=10$

$4)A'+B+C+D=1000=8$

$5)A+B'+C'+D=0110=6$

$6)A+B'+C+D=0100=4$

$7)A+B+C'+D=0010=2$

$8)A+B+C+D=0000=0$

$9)A'+B+C'+D'=1011=11$

$10)A'+B+C+D'=1001=9$

$11)A+B'+C'+D'=0111=7$

$12)A+B+C'+D'=0011=3$

$13)A'+B'+C+D'=1101=13$

$14)A+B'+C+D'=0101=5$

$15)A+B+C+D'=0001=1$

$F(A,B,C,D)=\prod (1,2,3,4,5,6,7,8,9,,10,11,12,13,14)$

$F(A,B,C,D)=\sum (15)$

      $$ \textbf{(OR)}$$

Let's First Simplify it

$F=[D'+AB'+A'C+AC'D+A'C'D]'$

$F=[D'+AB'+A'C+C'D(A+A')]'$

$F=[D'+AB'+A'C+C'D.1]'$        $[A+A'=1]$

$F=[D'+AB'+A'C+C'D]'$

$F=(D')'.(AB')'.(A'C)'.(C'D)'$       [ Using Demorgan's  Law$: (A+B)=A'.B'$  $(or)$ $(A.B)'=A'+B' ]$

$F=(D).(A'+B).(A+C').(C+D')$    [Again using Demorgan's law ]

$F=(A'D+BD).(AC+AD'+C'C+C'D')$   [Simple multiply]

$F=(A'D+BD).(AC+AD'+0+C'D')$     $[C.C'=0]$

$F=(A'D+BD).(AC+AD'+C'D')$

$F=(A'D.AC+A'D.AD'+A'D.C'D'+BD.AC+BD.AD'+BD.C'D')$

$F=ABCD$       $[A.A'=0,D.D'=0]$

    $$\textbf(OR)$$

Let $f(A,B,C,D) = \bigg[D'+AB'+A'C+AC'D+A'C'D\bigg]'$

$\implies \bigg[D'+AB'+A'C+C'D\bigg]'$

$\implies \bigg[D'+C'+ AB' + A'C\bigg]'$

$\implies \bigg[D'+C'+A'+ AB'\bigg]'$

$\implies \bigg[D'+C'+A'+ B'\bigg]'$​​​​​​​

$\implies ABCD$​​​​​​​

So the number of min-terms$=1$

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