1. Prove the identity of the boolean equation ab + bc'd + a'bc + c'd = b +c'd using algebraic equations.

2. Simplify the logic functions xyz + x'y + xyz'x' + xy + xz' +xy'z' using boolean algebra rules

 

i have try the first question. Is it correct??? any simpler way.. i mean short way.. this is long..

RHS
b +c'd
= b (1+a) + c'd
= b + ab + c'd
= (a+a')b + ab + c'd
= ab + a'b + ab + c'd
=ab + a'b (c+c') + ab + c'd
=ab + a'bc + a'bc' + ab + c'd
=ab + a'bc + b (a'c' +a) + c'd
=ab + a'bc + b(c' (a'+a) + c'd
=ab + a'bc + b(c') + c'd
=ab + a'bc + bc' + c'd
=ab + a'bc + bc'(d+d') + c'd
=ab + a'bc + bc'd + bc'd' + c'd
=ab + a'bc + c'd(b+1)+ bc'd'
=ab + b'c'd' + a'bc +c'd

 

The first exercise is wrong and its two parts aren't equal. The correct exercise should be ab+bc'd+a'b+c'd=b+c'd. You can work this out easilly if you group first the c'd factors together.

Try an attempt for the second one too. Remember that xx'=0 and (x+x')=1.