Step
*
2
1
1
of Lemma
p-fun-exp-compose
1. T : Type
2. n : ℤ
3. 0 < n
4. ∀[h,f:T ⟶ (T + Top)]. (f^n - 1 o h = primrec(n - 1;h;λi,g. f o g) ∈ (T ⟶ (T + Top)))
5. h : T ⟶ (T + Top)
6. f : T ⟶ (T + Top)
⊢ f o primrec(n - 1;p-id();λi,g. f o g) o h = f o primrec(n - 1;h;λi,g. f o g) ∈ (T ⟶ (T + Top))
BY
{ (Subst primrec(1 + (n - 1);p-id();λi,g. f o g) = f o primrec(n - 1;p-id();λi,g. f o g) ∈ (T ⟶ (T + Top)) 0⋅ THEN Auto\000C) }
1
.....equality.....
1. T : Type
2. n : ℤ
3. 0 < n
4. ∀[h,f:T ⟶ (T + Top)]. (f^n - 1 o h = primrec(n - 1;h;λi,g. f o g) ∈ (T ⟶ (T + Top)))
5. h : T ⟶ (T + Top)
6. f : T ⟶ (T + Top)
⊢ primrec(1 + (n - 1);p-id();λi,g. f o g) = f o primrec(n - 1;p-id();λi,g. f o g) ∈ (T ⟶ (T + Top))
2
1. T : Type
2. n : ℤ
3. 0 < n
4. ∀[h,f:T ⟶ (T + Top)]. (f^n - 1 o h = primrec(n - 1;h;λi,g. f o g) ∈ (T ⟶ (T + Top)))
5. h : T ⟶ (T + Top)
6. f : T ⟶ (T + Top)
⊢ f o primrec(n - 1;p-id();λi,g. f o g) o h = f o primrec(n - 1;h;λi,g. f o g) ∈ (T ⟶ (T + Top))
Latex:
Latex:
1. T : Type
2. n : \mBbbZ{}
3. 0 < n
4. \mforall{}[h,f:T {}\mrightarrow{} (T + Top)]. (f\^{}n - 1 o h = primrec(n - 1;h;\mlambda{}i,g. f o g))
5. h : T {}\mrightarrow{} (T + Top)
6. f : T {}\mrightarrow{} (T + Top)
\mvdash{} f o primrec(n - 1;p-id();\mlambda{}i,g. f o g) o h = f o primrec(n - 1;h;\mlambda{}i,g. f o g)
By
Latex:
(Subst primrec(1 + (n - 1);p-id();\mlambda{}i,g. f o g) = f o primrec(n - 1;p-id();\mlambda{}i,g. f o g) 0\mcdot{} THEN Auto)
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