A, B, C, and D are boolean variables, meaning that each takes the value
"true" or "false". More complex expressions have value "true" or "false"
depending on the values of these variables, so for example A'BD' is true if
A is false, B is true, and D is false, and C is either true or false.
To say that F = G, where F and G are complex expressions, means that, no
matter what values the boolean variables have, the value of F is the same
as the value of G (that is, F and G are either both true or both false).
So, for example, we have
AB + AC = A(B+C)
because if A is true and either B or C is true, then both sides are true,
and in any other case both sides are false---there's no way to assign
values to A, B, C such that the two side come out differently.
As previous posters have mentioned, you can always prove (or disprove!) an
equality by going through all possible assignments of "true" and "false" to
the variables. The goal seems to be to prove "by theorems", that is, using
operations previously proven true. As you say, we can either manipulate
one side of the equation until it has the form of the other side, or we
could manipulate both sides and get them into a common form.
In this case, the first thing to notice is that the right-hand side (RHS)
is a copy of the LHS with some extra stuff tacked onto the end. For this
reason it's simplest to manipulate just the RHS to get rid of the
surplussage! We can start with this basic theorem:
if Q is true whenever P is true, then Q = Q + P
We can prove this theorem by systematically considering all possibilities
for P and Q. Or look at it this way: If Q is true, then both sides of the
equation are true. And if Q is false, then P must be false (since, by
assumption, if P were true Q would be true) hence both sides are false.
Given this theorem we prove:
XY + X'Z = XY + X'Z + YZ
Proof: Let P be YZ and let Q be XY+X'Z. Suppose P is true, that is, Y and
Z are both true. But if Y and Z are both true, then XY+X'Z must be true.
(Reason: If X is true then, since Y is true, XY is true. And if X is
false then, since Z is true, X'Z is true. So in either case XY+X'Z is
true.) So we've shown that Q is true whenever P is true, hence by the
previous theorem Q = Q + P, which in this case is what we wanted to prove.
Now we can prove
A'D' + AC' = A'D' + AC' + C'D'
This is just the same as the previous theorem, putting A for X, D' for Y,
and C' for Z.
This last theorem implies
B(A'D' + AC') = B(A'D' + AC' + C'D')
which is the same as
A'BD' + ABC' = A'BD' + ABC' + BC'D'
Given this, we can take the RHS of the original and substitute A'BD' + ABC'
for A'BD' + ABC' + BC'D', which is to say, we can drop the term BC'D'.
With steps similar to the above we can prove these two theorems:
A'BD' + BCD = A'BD' + BCD + A'BC
BCD + ABC' = BCD + ABC' + ABD
which permit us to drop the final two terms of the RHS of the original,
completing the proof.
As far as I understand it correctly and $+$ means "or"; $\cdot$ means "and"; $\bar{x}$ means negation, it can surely be reduced. Btw, there is a mistake, because $(1)$ is not true if $x\bar y\bar z=1$, i.e., $(x,y,z)=(1,0,0)$, whereas $(2)$ obviously contains all $8$ cases and is therefore always true.
A general method to reduce such expressions is to "redistribute" by the following equivalent operations:
- $1=B+1=B+\bar B$ (maximality of $1$ and the tautology $B$ or $B$ negation);
- $A=AB+A\bar B$ for any $A,B$ (distributive law);
- $\overline{A+B}=\bar A\bar B$ and $A+B=\overline{\bar A\bar B}$ (de Morgan's laws).
- $\bar{\bar A}=A$ (negation is involution).
Using these we reduce $(1)$ to $\overline{x\bar y \bar z}$. There're surely algorithms for this (I don't know them), but it is similar to simplification of any formula, it takes some practice.
Best Answer
$$\begin{array}{ccc|cc} x & y & z & a & b \\ \hline 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0\end{array}$$ $$\implies \begin{cases}a= \overline x \cdot\overline y\cdot \overline z\\ b= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ a+b = \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z = b \\ a\cdot b = \overline x \cdot\overline y\cdot \overline z=a\end{cases}$$
Let's try to simplify $b$ a little:
$$\begin{align}b &= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x \cdot z + x \cdot \overline z) \\ &= \overline y \cdot [\overline x\cdot (\overline z + z) + x \cdot \overline z] \\ &= \overline y \cdot (\overline x + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot (\overline z+1) + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x + x\cdot \overline z) \\ &= \overline y \cdot ((\overline x +x)\cdot \overline z+\overline x) \\ &= \overline y \cdot (\overline z+\overline x) \\ &= \overline y \cdot \overline {z\cdot x}\end{align}$$
There's probably a slicker way of reaching that result, but it's been a while since I took electronics. But this way works! ;)