continuity while the second lemma proves a generalisation related to weak
anticontinuity. Lemma 6 will be applied immediately in the proof of lemma
8 below, but the application of lemma 7 will have to wait a bit, until we come
to the proof of Scott’s Induction Rule.
Lemma 6 Let Aα be continuous in ξ1 , . . . , ξn, then
[ V (A, α[RJ7ξι,..., Rn∙∕ξn]) = V (A,α[RT∕ξι, R£∕ξn])
m
Proof. Repeated application of continuity gives us that
[ ... [ V (A, a[R™ ∕ξι,..., Rnmn ∕ξn]) = V (A, a[Rω ∕ξι,..., R ∕ξn])
m1 mn
That the lefthand side of this equation is equal to
[ V (A, a[R1m ∕ξ1,...,Rnm∕ξn])
m
follows by the monotonicity of A in each of the variables ξ1,..., ξn. □
Lemma 7 Let Aα be continuous in ξ1 , . . . , ξn , while B is weakly anticontin-
uous in ζα, and ξ1 , . . . , ξn are not free in B. For each k (1 ≤ k ≤ n), let
R0k ⊆ R1k ⊆ . . . ⊆ Rkm ⊆ . . . be a chain, then
\ V([A∕ζ]B, a[R1m∕ξ1,..., Rnm∕ξn]) ⊆V([A∕ζ]B,a[R1ω∕ξ1,...,Rnω∕ξn])
m
Proof. For each m ∈ N let Sm = V(A, a[R1m ∕ξ1 , . . . , Rnm ∕ξn ]). From the
monotonicity of A in each ξk we obtain that S0 ⊆ S1 ⊆ . . . ⊆ Sm ⊆ ....
Hence, by the usual Substitution Theorem and the weak anticontinuity of B
in ζ,
\ V ([A∕ζ]B, a[RT∕ξι,..., Rn∙∕ξJ) = \ V (B, a[S m∕ζ ]) ⊆ V (B, a[Sω ∕ζ ])
mm
From Lemma 6 it follows that Sω = V(A, a [R1ω∕ξ1, . . . , Rnω∕ξn ]) for each m,
so that
V(B, a[Sω∕ζ]) =V([A∕ζ]B,a[R1ω∕ξ1,...,Rnω∕ξn])
□
18
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