Using the tabular method of simplification, find all equally minimal solutions for the function below.

$Z=f(A,B,C,D)=(1,4,5,10,12,14)$

Simplify the following Boolean function using K$-$Map simplification for three variables.

$F(a,b,c)=(0,1,2,5,7)$

$F(w,x,y,z)=,(0,2,4,5,6,7,8,10,13,15)$

For the function:

$,F=B+AB+AC+$ABC$+AC+C$

Construct a Karnaugh map and use it to find a minimum sum of products expression for $F$ and draw a logic circuit for the

minimized sum of products expression for f using AND, OR and NOT gates.

Map the function having four variables in Karnaugh's map. The function is

$,F(A,B,C,D)=(2,6,10,14).$

Simplify following with K$-$maps:

$-,F(A,B)=(1,2,3)$

$-,F(A,B,C)\frac{?}{b}=ar(AB)\frac{+}{b}ar(ABC)+A\frac{B}{b}ar(C)+\frac{A}{b}ar(BC)$

$-,F(A,B,C)=$ABC$+AB$

$-,F(A,B,C)\frac{?}{b}=ar(AC)\frac{+}{b}ar(B)C$

$-,F(A,B,C,D)\frac{?}{b}=ar(ABC)+\frac{A}{b}ar(CD)+\frac{A}{b}ar(B)+AB\frac{C}{b}ar(D)\frac{+}{b}ar(ABC)$

$-,F(A,B,C,D)=\frac{A}{b}ar(BD)\frac{+}{b}ar(ABD)+$ABD$+BCD+$ABCD

$-,F(A,B,C,D)\frac{?}{b}=ar(ABCD)\frac{+}{b}ar\frac{AB}{b}ar(D)+\frac{A}{b}ar(BCD)+\frac{A}{b}ar(B)\frac{C}{b}ar(D)$

$-,F(A,B,C,D)=(1,3,4,6,9,11)$

$-,F(A,B,C,D)\frac{?}{b}=ar()(2,3,6,7,12,13,14,15)$

$-,F(A,B,C,D)\frac{?}{b}=ar(AB)CD\frac{+}{b}ar(A)BCD+$ABCD$+\frac{A}{b}ar(B)CD+\frac{A}{b}ar(B)CD+A\frac{B}{b}ar(CD)+A\frac{B}{b}ar(C)D+AB\frac{C}{b}ar(D)$