There is no infinite sequence $(t_i)_{i\in\N}$ such that $t_0\in\mbb{U}$ and for all $i\in\N$, $t_i\surterm_{acc}t_{i+1}$.

\end{lem}

\begin{proof}

Let $f\,\bar u\in\mbb{U}$.

We want to show there is no infinite sequence $(t_i)_{i\in\N}$ such that $t_0=f\,\bar u$ and for all $i\in\N$, $t_i\surterm_{acc}t_{i+1}$.

Let us assume that there is such an infinite sequence.

First note that every $t_i$ is headed by an element of $\CObj$, otherwise there is no $t$ such that $t_i\surterm_{acc}t$.

Hence, among every infinite sequence,

let us choose one such that $t_0=f\,\bar u$

where $\tau(f)=\Pi\overline{(x:T)}.C\,\bar v$

and for all $C'\prec C$, there is no infinite sequence starting by a constructor of $C'$.

We will prove that for every $i$, $t_i=t'_i\,\bar v$

where $t'_0=t_0$, for all $i$, $t'_i\surterm t'_{i+1}$ and $t'_i$ is headed by a constructor of a type equivalent to $C$.

Let $i$ be such that $t_i$ has this property.

Hence $t_i=t'_i\,\bar v$.

Since $t_i$ is headed by a constructor,

$t'_i$ too, $t'_i=c\,\bar u$.

Since $t'_i\,\bar v\surterm_{acc}t_{i+1}$, two options:

\begin{itemize}

\item

If $f\notin\CObj$, then there is no $t$ such that $f\,\bar u\surterm_{acc}t$.

\item

So, let us consider $f\in\CObj$.

Then $\tau(f)=\Pi\overline{(x:T)}.C\,\bar v$.

By induction on $\succ$,

let's assume that for all $C'\prec C$,

for all terms headed by a constructor of $C'$,

there is no infinite sequence with the expected property.

There is a $j\in\Acc(f)$ such that $t_1=u_j\,\bar w$.

$u_j\in\interp{T_j}$, hence:

\begin{itemize}

\item

if $T_j$ is not a product and $u_j$ is headed by a constructor,

the constructor is fully applied, so $\valabs{\bar w}=0$.

\item

if $T_j$ is a product,

then by definition of accessibility, $T_j\in\Froz_{\<C}$.

So all the element of $\bar w$ headed by a constructor

are headed by a constructor of a type constant $C'\prec C$.

So if we choose to access an element of $\bar w$ we can apply the induction hypothesis on $\prec$.

\end{itemize}

We see that $\surterm_{acc}$ is well-founded on $\mbb{U}$

because $\surterm$ is well-founded and we can access finitely many invented terms, by well-foundedness of $\succ$.

\qedhere

\item

if $t_{i+1}=u_j\,\bar x$, then $u_j\subterm t'_{i}$ hence there is a $t'_{i+1}$ (namely $u_j$) such that $t_{i+1}=t'_{i+1}\,\bar x$ such that $t'_i\surterm t'_{i+1}$.

By definition of accessibility, if $\tau(c)=\Pi\overline{(x:T)}.C'\,\bar w$ with $C'\sim C$,

$T_j\in\Froz_{\<C}$.

In particular, $T_j=\Pi\overline{(y:V)}.C''\,\bar u'$,

with $C''\preceq C$.

Hence $u_j$ is headed by a constructor of $C''$.

\begin{itemize}

\item

If $C''\sim C$, we can conclude the induction,

\item

If $C''\prec C$, the minimality hypothesis is violated.

\end{itemize}

\item

if $t_{i+1}=v_j\,\bar x$, then we contradict the minimality hypothesis.

Indeed, $t_{i-1}=c_{i-1}\,\bar w$, with $c_{i-1}\in\CObj$,