Step
*
of Lemma
very-dep-fun-eta
No Annotations
∀[A,B:Type]. ∀[C:A ⟶ B ⟶ Type]. ∀[f:very-dep-fun(A;B;a,b.C[a;b])]. (f = (λL,b. (f L b)) ∈ very-dep-fun(A;B;a,b.C[a;b]\000C))
BY
{ (Auto
THEN All (Unfold `very-dep-fun`)
THEN (Enough to prove ∀n:ℕ. (f = (λL,b. (f L b)) ∈ vdf(A;B;a,b.C[a;b];n))
Because (EqTypeCD
THEN Auto
THEN ((Decide ⌜0 ≤ n⌝⋅ THENA Auto)
THENL [((D -3 With ⌜n⌝ THENA Auto) THEN D -4 With ⌜n⌝ THEN Auto)
; ((D -4 With ⌜n⌝ THENA Auto)
THEN (Unfold `vdf` -1 THEN (OReduce (-1) THENA Auto))
THEN Unfold `vdf` 0
THEN OReduce 0
THEN Auto
THEN GenConcl ⌜f = F ∈ ({L:(a:A × b:B × C[a;b]) List| ||L|| = 0 ∈ ℤ} ⟶ B ⟶ A)⌝⋅
THEN Auto)]
)))) }
1
1. A : Type
2. B : Type
3. C : A ⟶ B ⟶ Type
4. f : ⋂n:ℤ. vdf(A;B;a,b.C[a;b];n)
⊢ ∀n:ℕ. (f = (λL,b. (f L b)) ∈ vdf(A;B;a,b.C[a;b];n))
Latex:
Latex:
No Annotations
\mforall{}[A,B:Type]. \mforall{}[C:A {}\mrightarrow{} B {}\mrightarrow{} Type]. \mforall{}[f:very-dep-fun(A;B;a,b.C[a;b])]. (f = (\mlambda{}L,b. (f L b)))
By
Latex:
(Auto
THEN All (Unfold `very-dep-fun`)
THEN (Enough to prove \mforall{}n:\mBbbN{}. (f = (\mlambda{}L,b. (f L b)))
Because (EqTypeCD
THEN Auto
THEN ((Decide \mkleeneopen{}0 \mleq{} n\mkleeneclose{}\mcdot{} THENA Auto)
THENL [((D -3 With \mkleeneopen{}n\mkleeneclose{} THENA Auto) THEN D -4 With \mkleeneopen{}n\mkleeneclose{} THEN Auto)
; ((D -4 With \mkleeneopen{}n\mkleeneclose{} THENA Auto)
THEN (Unfold `vdf` -1 THEN (OReduce (-1) THENA Auto))
THEN Unfold `vdf` 0
THEN OReduce 0
THEN Auto
THEN GenConcl \mkleeneopen{}f = F\mkleeneclose{}\mcdot{}
THEN Auto)]
))))
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