Simplifying $(\neg x\land \neg y \land \neg z) \lor (\neg x\land \neg y \land z) \lor (x\land \neg y \land z) \lor ( x\land y \land z)$ I'm looking at this logical formula:
$(\neg x\land \neg y \land \neg z) \lor (\neg x\land \neg y \land  z) \lor (x\land \neg y \land z) \lor ( x\land  y \land  z)$
Asked to simplify it as much as possible, and I did something like this:
$((\neg x \land \neg y)\land(\neg z \lor z)) \lor ((x \land z)\land(y \land \neg y)) \equiv (\neg x \land \neg y) \lor (x \land z)$
The tutors book lists my solution along with an alternative one with this as the first step:
$(1):(x \land \neg x) \lor (\neg x \land \neg y) \lor (x \land z) \lor (\neg y \land z)$
along with some more manipulation to reach:
$(\neg x \lor z) \land (x \lor \neg y)$
Which left me puzzled as this is one step, and can't figure out which rules they used to get to $(1)$ (I presume it's some standard rule). Any ideas?
 A: Make a 3-dimensional truth table. Here, I'll just make 2 two-dimensional truth tables:
$z$:
$$\begin{array}{ccc}
&x&\text{not }x\\
y&\checkmark\\
\text{not }y&\checkmark&\checkmark
\end{array}$$
not $z$:
$$\begin{array}{ccc}
&x&\text{not }x\\
y&\\
\text{not }y&&\checkmark
\end{array}$$
It seems simplest to me to group the first two checkmarks as $x\land z$, and the other two as $\lnot y\land\lnot x$, so that all together makes $$(x\land z)\lor(\lnot y\land\lnot x)$$
However we can view the collection of checkmarks as an intersection instead of a union. For the language I will use, view the entire three-dimensional table as a cube, with the "$z$" table as its top face, and the "not $z$" table as its bottom face. The checkmarks are the intersection of 


*

*the top face unioned with the right face

*the left face unioned with the front face


This translates to the book's answer: $(z\lor\lnot x)\land(x\lor\lnot y)$
A: In (1) the term : $(¬x∧¬y)$ comes from :

$(¬x∧¬y∧¬z)∨(¬x∧¬y∧z) \equiv [(¬x∧¬y) \land (z \lor \lnot z)] \equiv [(¬x∧¬y) \land T] \equiv (¬x∧¬y).$


The term : $(x \land z)$ comes from :

$(x∧¬y∧z)∨(x∧y∧z) \equiv [(x \land z) \land (y \lor \lnot y)] \equiv [(x \land z) \land T] \equiv (x \land z).$


Applying idempotency : $a \lor a \equiv a$, we can "duplicate" the 2nd and 3rd disjuncts :

$(¬x∧¬y∧z)∨(x∧¬y∧z) \equiv (¬y∧z) \lor (x \land \lnot x).$


Thus, the original formula is equivalent to :


$(¬x∧¬y) \lor (x \land z) \lor (¬y∧z) \lor (x \land \lnot x).$



Now we have :
$(x∧z)∨(¬y∧z) \equiv (x∨¬y)∧z$
$(x∧¬x)∨(¬x∧¬y) \equiv (x∨¬y)∧¬x$
and putting all together :

$[(x∨¬y)∧z] ∨ [(x∨¬y)∧¬x] \equiv [(x∨¬y)∧(¬x∨z)]$.

