# Boolean Expression Orders of Operation

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???

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When you go from $(w'yz+x')$ to $(w+y'+z')x$ (just before "and given that"), aren't you missing a negation? $$(w'yz+x') = \Bigl( (w'yz)'x\Bigr)' = \Bigl((w+y'+z')x\Bigr)'.$$ – Arturo Magidin Apr 20 '11 at 20:59
I forgot to make it F' = and also take the complement while I was copying it over, yes. I edited it in thank you! – Xittenn Apr 20 '11 at 21:20
@PrettyFlower: Now your solution is correct, and your only problem seems to be that you were given an incorrect "appropriate solution" :-) – joriki Apr 20 '11 at 21:25
I hadn't written in all the steps properly. Given your correction(edited in), which I omitted in my original solution, I still get the same final answer. :/ – Xittenn Apr 20 '11 at 21:38
Very sorry about my attention to detail on this post! – Xittenn Apr 20 '11 at 21:43

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.

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(as per the last) or the right hand side were complemented. – Arturo Magidin Apr 20 '11 at 21:02
@Arturo: True :-) – joriki Apr 20 '11 at 21:03