I am trying reduce this DNF function to minimal variables. $f(a,b,c,d)=(ac’+c)(a’bc+d’)+(cd’+b)(cd’+c)+abd’+abc’d$
I have reduced to $ac'd+bc+cd'+abc'$ but I know it can be reduced down to $ab +ad'+bc+ cd'$
I am stuck and don't know where to continue from to get final solution.
Actually, you can't reduce it to $ab + ad' + bc + cd'$. Take $a = 1, b = c = d = 0$. Your reduced expression equals $0$, while that expression equals $1$ (because of the $ad'$ term).
However, using the absorption law $A + AB = A$ you can reduce it a little: $$ac'd + bc + cd' + abc' = ac'd + cd' + (bc + abc) + abc'$$ $$= ac'd + cd' + bc + ab(c+c') = ac'd + cd' + bc + ab.$$
Updated: I think you've made a mistake while reducing $f(a, b, c, d)$. Using absorption, complementation $A = xA + x'A$ and $x + x'y = x + y$ we have $$(ac'+c)(a'bc+d') + (cd'+b)(cd'+c) + abd' + abc'd = ac'd' + a'bc + cd' + bc + abd' +abc'd = ac'd' + cd' + bc + abcd' + abc'd' + abc'd = ac'd' + cd' + bc + abc'd = ac'd' + cd' + b(c + ac'd) = ac'd' + cd' + b(c+ad) = ac'd' + cd' + bc + abd = (ac' + c)d' + bc + abd = (a + c)d' + bc + abd = ad' + cd' + bc + abd = cd' + bc + a(d' + bd) = cd' + bc + a(d' + b) = ab + ad' + bc + cd'.$$
So this minimal form is correct, but you can't obtain it from your reduced expression since you've made a mistake.