Step
*
1
3
of Lemma
unsquashed-BIM-implies-unsquashed-weak-continuity-old
1. ∀B,Q:n:ℕ ⟶ (ℕn ⟶ ℕ) ⟶ ℙ.
((∀n:ℕ. ∀s:ℕn ⟶ ℕ. ((∀m:ℕ. Q[n + 1;s.m@n])
⇒ Q[n;s]))
⇒ (∀f:ℕ ⟶ ℕ. ⇃(∃n:ℕ. Q[n;f]))
⇒ (∀n:ℕ. ∀s:ℕn ⟶ ℕ. ∀m:ℕ. (B[n;s]
⇒ B[n + 1;s.m@n]))
⇒ (∀n:ℕ. ∀s:ℕn ⟶ ℕ. (B[n;s]
⇒ Q[n;s]))
⇒ Q[0;λx.⊥])
2. F : (ℕ ⟶ ℕ) ⟶ ℕ
3. a : ℕ ⟶ ℕ
4. n : ℕ
5. s : ℕn ⟶ ℕ
6. m : ℕ
7. ∀b:ℕ ⟶ ℕ. ((∀i:ℕn. ((rep-seq-from(s;n;a) i) = (b i) ∈ ℕ))
⇒ ((F rep-seq-from(s;n;a)) = (F b) ∈ ℕ))
⊢ ∀b:ℕ ⟶ ℕ
((∀i:ℕn + 1. ((rep-seq-from(s.m@n;n + 1;a) i) = (b i) ∈ ℕ))
⇒ ((F rep-seq-from(s.m@n;n + 1;a)) = (F b) ∈ ℕ))
BY
{ (UnivCD THENA Auto) }
1
1. ∀B,Q:n:ℕ ⟶ (ℕn ⟶ ℕ) ⟶ ℙ.
((∀n:ℕ. ∀s:ℕn ⟶ ℕ. ((∀m:ℕ. Q[n + 1;s.m@n])
⇒ Q[n;s]))
⇒ (∀f:ℕ ⟶ ℕ. ⇃(∃n:ℕ. Q[n;f]))
⇒ (∀n:ℕ. ∀s:ℕn ⟶ ℕ. ∀m:ℕ. (B[n;s]
⇒ B[n + 1;s.m@n]))
⇒ (∀n:ℕ. ∀s:ℕn ⟶ ℕ. (B[n;s]
⇒ Q[n;s]))
⇒ Q[0;λx.⊥])
2. F : (ℕ ⟶ ℕ) ⟶ ℕ
3. a : ℕ ⟶ ℕ
4. n : ℕ
5. s : ℕn ⟶ ℕ
6. m : ℕ
7. ∀b:ℕ ⟶ ℕ. ((∀i:ℕn. ((rep-seq-from(s;n;a) i) = (b i) ∈ ℕ))
⇒ ((F rep-seq-from(s;n;a)) = (F b) ∈ ℕ))
8. b : ℕ ⟶ ℕ
9. ∀i:ℕn + 1. ((rep-seq-from(s.m@n;n + 1;a) i) = (b i) ∈ ℕ)
⊢ (F rep-seq-from(s.m@n;n + 1;a)) = (F b) ∈ ℕ
Latex:
Latex:
1. \mforall{}B,Q:n:\mBbbN{} {}\mrightarrow{} (\mBbbN{}n {}\mrightarrow{} \mBbbN{}) {}\mrightarrow{} \mBbbP{}.
((\mforall{}n:\mBbbN{}. \mforall{}s:\mBbbN{}n {}\mrightarrow{} \mBbbN{}. ((\mforall{}m:\mBbbN{}. Q[n + 1;s.m@n]) {}\mRightarrow{} Q[n;s]))
{}\mRightarrow{} (\mforall{}f:\mBbbN{} {}\mrightarrow{} \mBbbN{}. \00D9(\mexists{}n:\mBbbN{}. Q[n;f]))
{}\mRightarrow{} (\mforall{}n:\mBbbN{}. \mforall{}s:\mBbbN{}n {}\mrightarrow{} \mBbbN{}. \mforall{}m:\mBbbN{}. (B[n;s] {}\mRightarrow{} B[n + 1;s.m@n]))
{}\mRightarrow{} (\mforall{}n:\mBbbN{}. \mforall{}s:\mBbbN{}n {}\mrightarrow{} \mBbbN{}. (B[n;s] {}\mRightarrow{} Q[n;s]))
{}\mRightarrow{} Q[0;\mlambda{}x.\mbot{}])
2. F : (\mBbbN{} {}\mrightarrow{} \mBbbN{}) {}\mrightarrow{} \mBbbN{}
3. a : \mBbbN{} {}\mrightarrow{} \mBbbN{}
4. n : \mBbbN{}
5. s : \mBbbN{}n {}\mrightarrow{} \mBbbN{}
6. m : \mBbbN{}
7. \mforall{}b:\mBbbN{} {}\mrightarrow{} \mBbbN{}. ((\mforall{}i:\mBbbN{}n. ((rep-seq-from(s;n;a) i) = (b i))) {}\mRightarrow{} ((F rep-seq-from(s;n;a)) = (F b)))
\mvdash{} \mforall{}b:\mBbbN{} {}\mrightarrow{} \mBbbN{}
((\mforall{}i:\mBbbN{}n + 1. ((rep-seq-from(s.m@n;n + 1;a) i) = (b i)))
{}\mRightarrow{} ((F rep-seq-from(s.m@n;n + 1;a)) = (F b)))
By
Latex:
(UnivCD THENA Auto)
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