Simplifying Boolean Expression (De Morgan)

Question

my question is basically, have I done the simplification correct so far? Is the answer correct? What is that simplification step I'm missing? Could anyone possibly outline the simplification steps for me in this question as I'm slightly confused?

Answer 1

Your very first step is incorrect; you failed to apply DeMorgan's Rule(s): $$\begin{align} \overline{A +B} &= \overline{A}\cdot \overline B\\ \overline{A\cdot B} &= \overline A +\overline B \end{align}$$

The missing step you're likely referring to is one of two distributivity laws:

$$\begin{align} A\cdot(B + C) = (B+C)\cdot A &= (A \cdot B) +(A\cdot C)\;\quad(1)\\ A+(B\cdot C) = (B\cdot C)+ A & = (A+B)\cdot(A+C)\quad(2)\end{align}$$

Applying this knowledge to your problem, $$\overline{\overline{(A+B)} +B} = \overline{\overline{A+B}}\cdot \overline{B} \tag{By Demorgan's}$$

$$= (A+B)\cdot \overline{B}\tag{using double negation}$$ $$A\cdot \overline{B} + \underbrace{B\cdot\overline{B}}_{\large =\, 0}\tag{distributive law (1)} $$

So that we are left with $A\cdot \overline B$

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Alternatively, I'll use $\lnot (x)$ to mean $\overline{x}$: (NOT x), and use parentheses to indicate the scope of each negation>

$$\lnot(\lnot(A +B) +B) = \lnot\lnot (A + B)\cdot( \lnot B) \tag{DeMorgan's}$$

$=(A + B)\cdot (\lnot B))\tag{Double Negation}$

$$= A\cdot( \lnot B)+\underbrace{(B\cdot(\lnot B))}_{= 0}\tag{distributive law}$$

$$= A \cdot (\lnot B)\tag{$(\text{from}\;A\cdot(\lnot B))+ 0$}$$

Answer 2

@amWhy already pointed out the mistake in your first step, and correctly indicated that on line 3 you should have:

$(A \lor B) \land \neg B$

What I want to add to this is that this last line is such a common pattern that there is an equivalence rule for it:

Reduction

$P \land (\neg P \lor Q) \Leftrightarrow P \land Q$

(this makes sense, since for $\neg P \lor Q$ to be true in addition to $P$ being true means that $Q$ has to be true. In other words, the term $P$ makes the term $\neg P \lor Q$ reduce to just $Q$)

And of course, for any equivalence, we always have a dual equivalence by swapping the $\land$'s and $\lor$'s, so we also have:

$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$

Anyway, given this Reduction principle, you last step is really just one step:

$(A \lor B) \land \neg B$ (Reduction!)

$A \land \neg B$

So yes, remember this Reduction rule: it's real handy to have when doing Boolean Algebra!