Using DeMorgan's Law, write an expression for the complement of F if
F(w, x, y, z) = xyz'(y'z + x)' + (w'yz + x')
F' = (xyz'(y'z + x)' + (w'yz + x'))'
= (xyz'(y'z + x)')' * (w'yz + x')'
= ( (xyz')' + (y'z + x) ) * ( (w'yz)' * x )
= ( ( (xy)' + z ) + (y'z + x) ) * ( ( w + (yz)' ) * x )
= ( ( (x' + y') + z ) + (y'z + x) ) * ( ( w + ( y' + z') ) * x )
= x' + y' + z + (y'z + x)(w + y' + z')x
where this is the appropriate solution given.
But now if we first start like this:
F = xyz'(y'z + x)' + (w'yz + x')
= xyz'((y'z)'x') + (w'yz + x')
= xyz'(((y')' + z')x') + (w'yz + x')
= xyz'((y + z')x') + (w'yz + x')
= xyz'(x'y + x'z') + (w'yz + x')
= xy(x'yz' + x'z') + (w'yz + x')
= x(x'yz' + x'yz') + (w'yz + x')
= xx'yz' + xx'yz' + (w'yz + x')
= xx'yz' + (w'yz + x')
= 0(yz') + (w'yz + x')
F'= (w'yz + x')'
= (w + y' + z')x
and given that
x' + y' + z + ((y'z + x)((w + y' + z')x))
= ((x' + y' + z) + (y'z + x))((x' + y' + z) + ((w + y' + z')x))
= (1 + y' + z + y'z) * ((x' + y' + z) + ((w + y' + z')x))
= (x' + y' + z) + ((w + y' + z')x)
= ((x' + y' + z) + (w + y' + z')) * ((x' + y' + z) + x)
= x' + y' + z + w + y' + z'
= 1
but
(w + y' + z')x != 1
then what have I done that causes the inequality???
Best Answer
Fortunately it is not within your powers to cause inequalities :-)
There are mistakes both in what you call the "appropriate solution" and in your alternative solution. That the appropriate solution can't be right follows from the fact that, as you correctly deduced, it reduces to $F'\equiv1$, whereas $F\not\equiv0$.
The mistake in the derivation of the appropriate solution is in the first step,
$$(xyz'(y'z + x)' + (w'yz + x'))'= ((xyz')') + ((y'z + x)(w'yz + x')')\;,$$
which should be
$$(xyz'(y'z + x)' + (w'yz + x'))'= ((xyz'(y'z + x)')'(w'yz + x')'\;.$$
The mistake in your alternative solution is the last step,
$$(w'yz + x')= (w + y' + z')x\;,$$
which would be a correct application of De Morgan's law if the left-hand side had been complemented.