I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, \ bx = b, \ cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
I think it could be done using a 27 state Markov chain process.
I propose using 27 states to represent our knowledge at any given time about the following three: $A$, $B$, $A \implies B$.
$[QQQ]$ means that we do not know whether any of the three are true or false.
$[QTF]$ means that we do not know whether $A$ is true or false, we know that B is true and we know that $A \implies B$ is false.
The states are therefore: $[QQQ]$,$[QQT]$,$[QQF]$,$[QTQ]$,$[QTT]$,$[QTF]$,$[QFQ]$,$[QFT]$,$[QFF]$,$[TQQ]$,$[TQT]$,$[TQF]$,$[TTQ]$,$[TTT]$,$[TTF]$,$[TFQ]$,$[TFT]$,$[TFF]$,$[FQQ]$,$[FQT]$,$[FQF]$,$[FTQ]$,$[FTT]$,$[FTF]$,$[FFQ]$,$[FFT]$,$[FFF]$.
It would be possible to create a transition matrix showing the conclusion that can be drawn from the state you are in at the moment.
So if you start in $[QQQ]$ you will always be in $[QQQ]$. There will be a lot of states like that, where no further information can be gained.
But if you are in $[TQT]$ then you must move to $[TTT]$. And once you are in $[TTT]$ you will stay there for higher powers of the matrix.
States are:
- $[QQQ]$
- $[QQT]$
- $[QQF]$
- $[QTQ]$
- $[QTT]$
- $[QTF]$
- $[QFQ]$
- $[QFT]$
- $[QFF]$
- $[TQQ]$
- $[TQT]$
- $[TQF]$
- $[TTQ]$
- $[TTT]$
- $[TTF]$
- $[TFQ]$
- $[TFT]$
- $[TFF]$
- $[FQQ]$
- $[FQT]$
- $[FQF]$
- $[FTQ]$
- $[FTT]$
- $[FTF]$
- $[FFQ]$
- $[FFT]$
- $[FFF]$
The $27 \times 27$ matrix $M$ has entries $M_{ij}=1$ if being in State $i$ will allow you to move to State $j$.
Best Answer
Here's a hint: In a Boolean algebra, $a \vee a = a \vee a \vee a$ for all elements $a$, but that doesn't mean that $O_B = a$ for all $a$. Thus I think you should look back at what kind of cancellation laws are valid.
http://en.wikipedia.org/wiki/Boolean_algebra