I'm new at learning boolean algebra and understand the basic laws. But now I'm trying to proof:
$x'z+ xyz + xy'z = z$
Can someone help me with this please ?
Thanks in advance.
[Math] Boolean Algebra Proof
logic
Related Solutions
The Concensus Theorem says:
$XY \lor Y'Z \lor XZ = XY \lor XZ$
So your exercise is really a particular application of this!
But let's assume the Concensus Theorem is not available for you to use.
Then you can do:
$(A \land B) \lor (\neg A \land C) \lor (B \land C) =$ (Adjacency)
$(A \land B) \lor (\neg A \land C) \lor (A \land B \land C) \lor (\neg A \land B \land C)=$ (Absorption x 2)
$(A \land B) \lor (\neg A \land C)$
This assumes:
Adjacency
$P = (P \land Q) \lor (P \land \neg Q)$
Absorption
$P = P \lor (P \land Q)$
A simple answer to your question is that Boolean algebra is not the same thing as 'ordinary' algebra (arithmetic). The intuition that connects $+$ with $\vee$ and $\times$ with $\wedge$ really is just intuition: the laws of addition and multiplication of numbers do not exactly coincide with the laws of the join and meet operations, and there's no reason to expect them to do so.
But there is a connection that can be made, which is if work with the integers modulo $2$, i.e. we only look at whether an integer is even or odd. In this case, it is true that $x+yz = (x+y)(x+z)$, since the quantity on the left is even (resp. odd) if and only if the quantity on the right is even (resp. odd).
A more mathematical answer for why $x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$ has to do with duality. Every boolean algebra $\mathbb{B}$ has a dual $\mathbb{B}^{\mathrm{op}}$, defined so that the meet of $\mathbb{B}^{\mathrm{op}}$ is the join of $\mathbb{B}$, the join of $\mathbb{B}^{\mathrm{op}}$ is the meet of $\mathbb{B}$, and the top and bottom elements are swapped. This means that any equation that holds in an arbitrary Boolean algebra must also be true in an arbitrary Boolean algebra if we swap $\wedge$ and $\vee$ (and $\top$ and $\bot$).
Best Answer
Are you still stuck? I'll use your terminology, which seems to be used in electrical engineering.
1) $x'z + xyz + xy'z$
2) $x'z + xz(y+y') $
Since $y+y' = 1$, and $xz1 = xz$ then
3) $x'z + xz$
4) $(x'+x)z$
5) $z$
Hope that helps.