= x +
\iota(x) - u = \alpha(x) + \beta(x)- u = \alpha(x) \boxplus
\beta(x).~$ Therefore the condition (3.1.3) from Definition 3.1 holds.

Let $ x,y \in V(u)$. Applying the properties of maps $ \alpha $ and $ \beta $ we have $ \alpha (x\boxdot y ) = \alpha (
(x-u)\odot(y-u) + u ) = \alpha ( (x-u)\odot(y-u)) + \alpha (u)
=$\
$=\alpha (x-u) + \alpha(u) = \alpha(x) = u $ and $ \beta (x\boxdot
y )= u $ and so $ x\boxdot y \in V(u).$ Hence the law $ \boxdot$ given by the relation (3.3) is well-defined.

If $ x, y, z \in V(u)$ then the following equality holds:\
$(e)~~~ x\boxdot ( y \boxplus z \boxplus ( \boxminus \beta(x))) =
( x\boxdot y ) \boxplus (  x\boxdot z  ) \boxplus ( \boxminus
x)).$\
Indeed, we have\
$(e.1)~~ x\boxdot ( y \boxplus z \boxplus ( \boxminus \beta(x))) =
x\boxdot ( y \boxplus z \boxplus ( \boxminus u)) = x\boxdot ( y
\boxplus z \boxplus u )=$\
$= x\boxdot ( y \boxplus z )= (x - u)\odot ( y \boxplus z - u ) +
u = (x - u)\odot ( (y - u) + (z - u )) + u. 