Question

Simplify each of the following two Boolean equations (using Boolean algebra, in particular consensus theorm).

ac'd' + ab'cd' + a'bcd' + bd + a'bc'd + abc

Answer 1

Given

ac'd' + ab'cd' + a'bcd' + bd + a'bc'd + abc

ac'd' + ab'cd' + a'bcd' + abc + bd + a'bc'd

ad'(c' + b'c) + bc(a'd' + a) + bd(1 + a'c')  

ad'[(c' + b') (c' + c)] + bc[(a' + a)(d' + a)] + bd[(1 + a')(1+c') ]  { By Distributive law P+QR = (P+Q)(P+R) }

ad'[(c' + b') (1)] + bc[(1)(d' + a)] + bd[(1)(1) ]  { We know that P+P'=1 &1+P'=1 }

ad'(c' + b') + bc(d' + a) + bd

(ac'd' + ab'd') + (bcd' + abc) + bd  { By Distributive law P(Q+R) = PQ+PR }

ac'd' + ab'd' + bcd' + abc + bd

ac'd' + ab'd' + bcd' + bd + abc

ac'd' + ab'd' + b(cd' + d) + abc   { By Distributive law PQ+PR = P(Q+R) }

ac'd' + ab'd' + b[(c + d)(d' + d)] + abc   { By Distributive law P+QR = (P+Q)(P+R) }

ac'd' + ab'd' + b[(c + d)(1)] + abc   { We know that P+P'=1 }

ac'd' + ab'd' + b(c + d) + abc

ac'd' + ab'd' + (bc + bd) + abc { By Distributive law P(Q+R) = PQ+PR }

ac'd' + ab'd' + bc + abc + bd

ac'd' + ab'd' + bc(1 + a) + bd    { By Distributive law PQ+PR = P(Q+R) }

ac'd' + ab'd' + bc(1) + bd   { We know that 1+P'=1 }

ac'd' + ab'd' + bc + bd

Which is Required Simplified Boolean equation