and $snd$ we obtain $f~n_p~g_p$ that we know is in relation $\R$ with $Q$
since $g_p \R P[\underline{n_p}/x]$.
+\item $ex\_intro$.
+ We must prove that
+ $$\lambda x:\N.\lambda f:\sem{P}.<x,f> \R \forall x.(P\to\exists x.P(x)$$
+ that leads to prove that foreach n
+ $\underline{ex\_into}~n \R (P\to\exists x.P(x))[\underline{n}/x]$.\\
+ Evaluating the substitution we have
+ $\underline{ex\_into}~n \R (P[\underline{n}/x]\to\exists x.P(x))$.\\
+ Again by definition of $\R$ we have to prove that given a
+ $m \R P[\underline{n}/x]$ then $\underline{ex\_into}~n~m \R \exists x.P(x)$.
+ Expanding the definition of $\underline{ex\_intro}$ we have
+ $<n,m> \R \exists x.P(x)$ that is true since $m \R P[\underline{n}/x]$.
+
\item $false\_ind$.
We have to prove that $\bot_{\sem{Q}} \R \bot \to Q$.
Trivial, since there is no $m \R \bot$.
Since there is no $n$ such that $0 = S n$ is true... \\
$\underline{discriminate}~n \R 0 = S~\underline{n} \to \bot$ for each n.
+\item $injS$.
+ We have to prove that for each $n_1$ and $n_2$\\
+ $\lambda \_:\N. \lambda \_:\N.\lambda \_:1.*~n_1~n_2 \R
+ (S(x)=S(y)\to x=y)[n_1/x][n_2/y]$.\\
+ Assuming $m \R S(n_1)=S(n_2)$ we have to prove
+
+
+
\end{itemize}
\begin{thebibliography}{}