$\nf{\tm\sigma}\EtaNeq\nf{\tmtwo\sigma}$.\r
Let $\tm \EtaEq \lambda\vec\vartwo.\, h_1\,\vec a$ and \r
$\tmtwo \EtaEq \lambda\vec\vartwo.\, h_2\,\vec b$,\r
- with $|\vec\vartwo| = \xK$.\r
+ with $|\vec\vartwo| = \xK$ {\color{red}[Capire perche' si puo' $\xK$-$\eta$-espandere qui]}.\r
By (course-of-value) induction on the size of $\tm$ and $\tmtwo$,\r
and by cases on the definition of eta-difference:\r
\begin{itemize}\r
\r
We now consider the other subcase $h_1=h_2=\var$, and the following subsubcases:\r
\begin{itemize}\r
- \item Both $|\vec a|,|\vec b|\leq i$: \TODO\r
- \item w.l.o.g. $|\vec a|>i$ and $|\vec b|\leq i$: use boh \TODO\r
- \item Both $|\vec a|,|\vec b|> i$: \TODO use the lemma above\r
+ \item $|\vec a|,|\vec b|\leq i$: \TODO\r
+ \item $|\vec a|>i$ and $|\vec b|\leq i$: use boh \TODO\r
+ \item $|\vec a|\leq i$ and $|\vec b|> i$: symmetric to the case above.\r
+ \item $|\vec a|,|\vec b|> i$: \TODO use the lemma above\r
plus another one. Reason on the numebr of arguments\r
$\Args{\nf{a_i\sigma\vec\alpha}}$ vs \r
$\Args{\nf{b_i\sigma\vec\alpha}}$.\r
\lambda\vec\vartwo.\,h_2\,\vec{(\nf{b\sigma})}$,\r
and we conclude by \ih{} since $a_n \EtaNeq b_n$ implies that \r
$\nf{a_n\sigma} \EtaNeq \nf{b_n\sigma}$.\r
- {\color{red}[Can we use the \ih{} here?]}\r
+ {\color{red}[Capire se possiamo davvero usare l'\ih{} qui]}\r
\r
We now consider the other subcase $h_1=h_2=\var$, and the following subsubcases:\r
\begin{itemize}\r