Proof that the $\dim (U+W)= \dim U + \dim W -\dim (U\cap W)$ The Book is Bosch Linear Algebra.
I have difficulties to understand how one deduces this statement from prior statements proved earlier:
Statement already proved:
Definition 1.
Let $U_1,....,U_r$ be linear subspaces of a $KVR$ $V$. The sum is defined as 
$$\sum_{i=1}^{r}U_i=\{\sum_{i=1}^{r}b_i;b_i\in U_i, i=1,...,r \}$$
Theorem 2Let $U=\sum_{i=1}^{r}U_i$ be a sum of subspaces of a $KVR$ $V$, the following statements are equivalent:
(i) If $b$ is written as $b=\sum_{i=1}^{r}$ then the $b_i$ are unique. (ii) If $\sum_{i=1}^{r}b_i=0$ then $b_i=0$ for all $i$ (iii) $U_p\cap\sum_{i\neq p}U_i=0$ for all $p=1,...,r$.
Definition 3 If the sum $U=\sum U_i$ fullfills one of the conditions above in Theorem 2 the sum is called direct.
Theorem 4Let $V$ be a $KVR$ $V$ and $U$ a subspace then there exissts a subspace $U'$ with $V=U+U'$ such that the sum is direct and for all such $U'$ the dimension formula:
$$\dim V= \dim U +\dim U'$$ holds

Theorem 5 $\dim (U+W)= \dim U + \dim W -\dim (U\cap W)$

There is a question about the proof, up to a certain point I understand it:
We suppose that $U$ and $U'$ are of finite dimension subspaces of a vectorspace in $V$.
$U+U'$ thus has a finite generating system and thus a finite basis. One chooses in accordance to Theorem 4 a complement $W$ of $U\cap U'$ and $W'$ of $U\cap U'$ such that 
$U=(U\cap U') + W$ where the sum is direct and $U'=(U \cap U') + W'$ where the sum is also direct.
Then 
$U+U'=(U\cap U')+W+W'$ and the sum (on the RHS) is also direct.
Let $a+b+b'=0$ $a\in$ intersection $b\in W$ and $b'\in W'$ then 
$b=-(a+b')\in (U\cap U')+W'=U'$ and because of $b\in W\subset U$. $b\in (U\cap 
 U')\cap W=0$
analogously $b'=0$ then also $a=0$. Therefore the sum is direct
And because of Theorem 4 we now can say
$\dim(U+U')=\dim(U\cap U') + \dim W + \dim W'$
This is a sum with $3$ summands, if it would be a sum with two summands then I would understand it. But when it is three sumands I don't know how you can directly deduce from Theorem 4 this dimension formula. You can prove it by induction but the author says this can be directly deduced from theorem $4$ and didn't even mention induction. 
I know there are many questions about this formula but I want to understand this specific proof of this formula. 
 A: There is a proposition which states that if $A,B,C$ are subspaces of $V$, then
$V=A\oplus B \oplus C \iff V=A+B+C \text{ and } A\cap(B+C)= B\cap(A+C)= C\cap(A+B)=\{0_V\}$
This means that if $V_1=A+B$ and $V_2=C$, then $V=V_1\oplus V_2$ and $\text{dim}V=\text{dim}V_1+\text{dim}V_2$. Moreover, if $$V_1=A\oplus B,$$ then $\text{dim}V_1=\text{dim}A+\text{dim}B$, hence 
\begin{equation}\tag{1}
\text{dim}V=\text{dim}V_1+\text{dim}V_2=\text{dim}A+\text{dim}B+\text{dim}C.
\end{equation}
Take now $V=U+U'$, $A=U\cap U'$, $B=W$, $C=W'$ and $V_1=A+B=(U\cap U')+W$. Then $V=U+U'=(U\cap U')\oplus W \oplus W'=A\oplus B \oplus C$. But since you have proved that $V_1=(U\cap U')+W=A+B$ is a direct sum, then $V_1=A\oplus B$, and so by using $(1)$ you have $$\text{dim}(U+U')=\text{dim}V=\text{dim}A+\text{dim}B+\text{dim}C=\text{dim}(U\cap U')+\text{dim}W+\text{dim}W'.$$
Is that what are you asking?
A: You can use Theorem 4 twice: Let $$A:=W \oplus W'.$$
Note that $A$ is also a vector space and that the sum is direct. So $\dim A = \dim W + \dim W'$ by Theorem 4.
Using Theorem 4 again, we see that $\dim (U+U')=\dim A +\dim(U\cap U')=\dim(U\cap U') +\dim W + \dim W'$.
