In this sub-case, we have by the inductive hypothesis, FV(G[u:=w]) = FV(G) and so we have, remembering that, from Lemma Sub.0, height(G[u:=w]) = height(G)=n
(G[u:=w])[x:=F] G[u:=w]
G
So,
by FV4. Moreover E[x:=F] E by RA3.
This proves our sub-case, since
x FV(G) and
x
FV(E)
Base Case n=0
Suppose for a given n, for all i n
we have
and hence x FV(F1),
x
FV(F2).
Then
by the inductive hypothesis.
We have
by S6. But, by FV4, the free variables of G are the free variables
of E with the possible addition of
u x. So x
FV(G), and we may apply our inductive hypothesis to obtain
But, by Lemma..., the free variables of G[u:=w] are the free variables of G, minus u, with the addition of w.
In this sub-case, by the inductive hypothesis,
By FV1. Thus x FV(G[u:=w])
or x=w.
= (FV(G[u:=w]) - w) FV(F)
= (((FV(G)-{u}) {w}) - {w})
FV(F)
On the other hand