Novice error while simplifying boolean expression with XOR

Question

I have the following expressions:

$x = a⊕b , y = c⊕d, z = x⊕y $

So $x = ab'+a'b, y = cd'+c'd $

$z = xy'+x'y = (ab'+a'b) * (cd'+c'd)' + (ab'+a'b)'*(cd'+c'd)$

If I take the first term : $(ab'+a'b) * (cd'+c'd)' = (De Morgan) = (ab'+a'b) * c'dcd'$

Yet, $c'dcd'$ is false, so the first term is false.

Where is my mistake?

Thank you

Answer 1

If I take the first term : $(ab'+a'b) * (cd'+c'd)' = (De Morgan) = (ab'+a'b) * c'dcd'$

Yet, $c'dcd'$ is false, so the first term is false.

Where is my mistake?

Your mistake is in the DeMorgan:

It is not true that $(cd'+c'd)'=c'dcd'$

Rather:

$(cd'+dc')' = (cd')'(dc')' = (c'+d)(d'+c)$

Answer 2

if you're in a boolean algebra

In a boolean algebra, $+$ denotes XOR already. $a \vee b$ corresponds to $a + b + ab$.

Besides (this may simplify your expressions nicely), the negation of $a$, which you write $a'$, is simply $1 + a$.

if you want to handle a boolean expression

There's a mistake in the de Morgan rule.

The first term is
$((a\wedge b')\vee (a'\wedge b)) \wedge ((c\wedge d')\vee (d'\wedge d))$
$ = [(a \wedge b') \wedge ((c\wedge d')\vee (c'\wedge d))] \vee [(a' \wedge b) \wedge ((c\wedge d')\vee (c'\wedge d))]$ $$=(a \wedge b' \wedge c \wedge d')\vee (a \wedge b' \wedge c' \wedge d)\vee (a' \wedge b \wedge c \wedge d') \vee (a' \wedge b \wedge c' \wedge d)$$