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.
It's a counting problem. It is based on the following:
Product Rule: If one even can occur in $k$ different ways, and a second event can occur in $m$ different ways, then the number of ways in which both events can occur is $km$.
That is, there are $k$ possible outcomes for the first event, and $m$ for the second, so there are $km$ total possible combinations of outcomes.
The truth table has one row for every possibility. Each variable has two possible outcomes: true or false. If there are $n$ different variables, then you have two possibilities for each, and the number of total combinations is the product of the number of possible outcomes for each. That amounts to a product of $2$ with itself, $n$ times, or $2^n$.
Inductively: one variable has $2$ possible outcomes. Assume that $k$ variables have $2^k$ possible outcomes. How many possible outcomes are there for $k+1$ variables? There are $2^k$ ways in which you can have outcomes in which the last variable is false, and another $2^k$ ways in which you can have outcomes in which the last variable is true. So there is a total of $2^k+2^k = 2(2^k) = 2^{k+1}$ different possible outcomes.
Best Answer
In this context, "unknown" means "depends on input", and you are right. An answer of "true" would mean "this expression is a tautology", i.e. returns true on all inputs, and likewise for "false" and the expression returning false on all inputs. Since both true and false values may be assumed by the given expression, the answer is "unknown".