will write $~G_{v}^{u} = \alpha^{-1}(u) \cap \beta^{-1}(v).~$

A groupoid $~( G, G_{0})$ is said to be [*transitive*]{}, if its anchor map is surjective.

[**Convention.**]{} (1) We write sometimes $~x y~$ for $~m(x,y)~$, if $ (x,y) \in G_{(2)}.$

(2) Whenever we write a product in a given groupoid, we are assuming that it is defined.$\b$

In the following proposition we summarize some basic rules of algebraic calculation in a Brandt groupoid obtained directly from definitions.

([@ivan02])

*In a groupoid $~( G, \alpha, \beta, m, \iota
, G_{0})~$ the following assertions hold :*

$(i)~~~\alpha(u)=\beta(u)=u,~~~ u\cdot u=u \quad \hbox{and}\quad
\iota (u)=u,~ \forall u\in G_0;$

$(ii)~\alpha \left( x\cdot y\right) =\alpha \left( x\right) \quad
\hbox{and}\quad \beta \left( x\cdot y\right) =\beta \left(
y\right) ,~ \forall  \left( x,y\right) \in G _{\left( 2\right) };$

$(iii)~\alpha \left( x^{-1}\right) =\beta \left( x\right) \quad
\hbox{and}\quad \beta \left( x^{-1}\right) =\alpha \left( x\right)
,~ \forall  x\in G;$

$(iv)~$ [**(cancellation law)**]{} If for $ x,y_1,y_2,z\in G $ we have $(x,y_1),(x,y_2),$ $(y_1,z),(y_2,z)\in G_{(2)},$ then: $$(a) \quad  x\cdot y_1=x\cdot y_2~~\Rightarrow~~ y_1=y_2;~~~
(b) \quad y_1\cdot z=y_2\cdot z~~\Rightarrow~~ y_1=y_2.$$

[*(v)*]{} For each $ x\in G$ we have $~(x^{-1})^{-1}=x.$

[*(vi)*]{} If $(x,y)\in G_{(2)},$ then $(y^{-1},x^{-1})\in G_{(2)}$ and the equality holds: $$(x\cdot y)^{-1}=y^{-1}\cdot x^{-1}.$$

[*(vii)*]{} For all $(x,y)\in G_{(2)}$, the 