Arithmetic in Lambert-W number system $\DeclareMathOperator{\sign}{sign}$
The principal branch of Lambert-W function is defined like so:
$$W(x) = f^{-1}(x)\\
  \text{where} f(x)=xe^x$$
So $W(e) = 1$, $W(2e^2)$ is 2, $W(3e^3)$ is 3 etc.
We define Lambert-W number system to represent a real number $X$:
$$X \rightarrow \{s=\sign(x),x=W(|x|)\}$$
For instance, $-2e^{2}$ is represented $\{-1,2\}$, and $1$ is represented as $\{1,W(1)\}$.

Additionally, we define double Lambert-W number system as this:
$$X \rightarrow \{s=\sign(x),x=W(W(|x|))\}$$
For instance, $(2e^{2})\cdot{e^{(2e^{2})}}$ is represented as $\{1,2\}$.
My question: How to add, subtract, multiply and divide numbers represented in these two systems?
 A: $\DeclareMathOperator{\sign}{sign}$
Let us consider $x \mapsto (\sign(x),W(|x|))$ and $y \mapsto (\sign(y),W(|y|))$.
This means that
$$ |x| = f(f^{-1}(|x|)) = f(W(|x|)) , $$
so
$$ x = \sign(x) |x| = \sign(x) f(W(|x|)) = \sign(x) W(|x|) e^{W(|x|)} . $$
At the same way,
$$ y = \sign(y) W(|y|) e^{W(|y|)} . $$
So, knowing the representation of $x$ and $y$ in the Lambert-W number system, we can reconstruct the value of $x$ and $y$.
Now assume that $\sign(x)=\sign(y)$.
$$ x+y = \sign(x) \left( W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} \right) , $$
which means that
$$ |x+y| = W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} . $$
The representation of $x+y$ in the Lambert-W number system will be then
$$ x+y \mapsto \left(\sign(x+y), W(|x+y|)\right) = \left(\sign(x+y), W\left( W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} \right) \right) . $$
We just proved that, if $(s_1,x_1)$ and $(s_2,x_2)$ are two numbers in the Lambert-W number system, if $s_1 = s_2$, we can write
$$ (s_1,x_1)+(s_2,x_2) = \left(s_1,W\left(x_1 e^{x_1} + x_2 e^{x_2} \right) \right) . $$
When $s_1 \neq s_2$, that is when $\sign(x) \neq \sign(y)$, you can do a similar calculation using the fact that the function $f$ is monotone, so if $x > y$, then $x e^x > y e^y$.
You can derive the other operations with similar arguments.
