So, using Church numerals, we define
$3 = {\lambda} f. {\lambda}x.f(f(f(x)))$,
and
$4 = {\lambda} f. {\lambda}x.f(f(f(f(x))))$.
We can then add with an expression like
$3\ g\ (4\ g\ z)$
And this reduces to:
$(g (g (g (g (g (g (g\ z)))))))$
... but why?
$g$ is a free variable in each expression, and my understanding is that you must ${\alpha}$-convert free standing variables in unrelated expressions. Shouldn't we instead end up with something like
$(g (g (g (g_2 (g_2 (g_2 (g_2\ z)))))))$?