Step
*
of Lemma
sub-free-dim-1
∀[K:CRng]. ∀[S,T:Type].
(∀s:T. ∀x:Point(sub-free-vs(K;S;T)). (↓∃k:|K|. (x = {<k, s>} ∈ Point(sub-free-vs(K;S;T))))) supposing
((∀x,y:T. (x = y ∈ T)) and
strong-subtype(T;S))
BY
{ (Auto
THEN RepUR ``vs-point sub-free-vs sub-vs mk-vs`` -1
THEN Fold `vs-point` (-1)
THEN D -1
THEN DupHyp (-1)
THEN RepeatFor 3 (D -1)
THEN RepUR ``vs-point sub-free-vs sub-vs mk-vs`` 0
THEN Fold `vs-point` 0) }
1
1. K : CRng
2. S : Type
3. T : Type
4. strong-subtype(T;S)
5. ∀x,y:T. (x = y ∈ T)
6. s : T
7. x : Point(free-vs(K;S))
8. fs-in-subtype(K;S;T;x)
9. b : basic-formal-sum(K;S)
10. x = b ∈ formal-sum(K;S)
11. bfs-predicate(K;S;p.snd(p) ∈ T;b)
⊢ ↓∃k:|K|. (x = {<k, s>} ∈ {f:Point(free-vs(K;S))| fs-in-subtype(K;S;T;f)} )
Latex:
Latex:
\mforall{}[K:CRng]. \mforall{}[S,T:Type].
(\mforall{}s:T. \mforall{}x:Point(sub-free-vs(K;S;T)). (\mdownarrow{}\mexists{}k:|K|. (x = \{<k, s>\}))) supposing
((\mforall{}x,y:T. (x = y)) and
strong-subtype(T;S))
By
Latex:
(Auto
THEN RepUR ``vs-point sub-free-vs sub-vs mk-vs`` -1
THEN Fold `vs-point` (-1)
THEN D -1
THEN DupHyp (-1)
THEN RepeatFor 3 (D -1)
THEN RepUR ``vs-point sub-free-vs sub-vs mk-vs`` 0
THEN Fold `vs-point` 0)
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