+ We assume that $m \R S(n_1)=S(n_2)$ and we have to show that
+ $\lambda \_:\N. \lambda \_:\N.\lambda \_:1.*~n_1~n_2~m$ that reduces to
+ $*$ is in relation $\R$ with $n_1=n_2$. Since in the standard model of
+ natural numbers $S(n_1)=S(n_2)$ implies $n_1=n_2$ we have that
+ $* \R n_1=n_2$.
+
+\item $plus\_O$.
+ Since in the standard model for natural numbers $0$ is the neutral element
+ for addition $\lambda \_:\N.\star \R \forall x.x + 0 = x$.
+
+\item $plus\_S$.
+ In the standard model of natural numbers the addition of two numbers is the
+ operation of counting the second starting from the first. So
+ $$\lambda \_:\N. \lambda \_:\N. \star \R \forall x,y.x+S(y)=S(x+y)$$.
+
+\item $times\_O$.
+ Since in the standard model for natural numbers $0$ is the absorbing element
+ for multiplication $\lambda \_:\N.\star \R \forall x.x \times 0 = x$.
+
+\item $times\_S$.
+ In the standard model of natural numbers the multiplications of two
+ numbers is the operation of adding the first to himself a number of times
+ equal to the second number. So
+ $$\lambda \_:\N. \lambda \_:\N. \star \R \forall x,y.x+S(y)=S(x+y)$$.