Can someone show me step by step how to simplify this boolean expression? I would like to learn how to handle this kind of simplifications:
$$ Y = \neg(D \wedge\neg E) \vee (\neg E \wedge D ) $$
I can apply boolean laws for the first steps, that should be:
- De Morgan's law : $\neg D \vee \neg\neg E \vee (\neg E \wedge D)$
- Elimination of double negation : $\neg D \vee E \vee (\neg E \wedge D)$
From here i don't know what to do. Except that final result is $\top$.
Distribute the result with the distribution law $A \vee ( B \wedge C ) = (A \vee B) \wedge (A \vee C )$. Letting $A = (\neg D \vee E)$, $B = \neg E$ and $C = D$ you obtain $$ (\neg D \vee E) \vee (\neg E \wedge D) = ((\neg D \vee E) \vee \neg E ) \wedge ((\neg D \vee E) \vee D ). $$
Obviously, $E \vee \neg E = \top$ and likewise for $D$. Besides $\top \vee A = \top$ for a term $A$. Therefore the expression reduces as follows.
$$ (\neg D \vee E \vee \neg E ) \wedge (\neg D \vee E \vee D ) = (\neg D \vee \top ) \wedge (\top \vee E ) = \top \wedge \top = \top $$
But you don't really need all that trouble. If you let $X = \neg E \wedge D$, your expression became $Y = \neg X \vee X$, which again equals $\top$.