Step
*
1
1
1
of Lemma
strong-continuity-implies4
1. F : (ℕ ⟶ ℕ) ⟶ ℕ
2. M : n:ℕ ⟶ (ℕn ⟶ ℕ) ⟶ (ℕn?)
3. ∀f:ℕ ⟶ ℕ. (↓∃n:ℕ. (((M n f) = (inl (F f)) ∈ (ℕ?)) ∧ (∀m:ℕ. ((↑isl(M m f))
⇒ ((M m f) = (inl (F f)) ∈ (ℕ?))))))
4. f : ℕ ⟶ ℕ
5. n : ℕ
6. (M n f) = (inl (F f)) ∈ (ℕ?)
7. ∀m:ℕ. ((↑isl(M m f))
⇒ ((M m f) = (inl (F f)) ∈ (ℕ?)))
⊢ <mu(λn.isl(M n f)), Ax> ∈ ∃n:ℕ. ((M n f) = (inl (F f)) ∈ (ℕ?))
BY
{ ((Assert isl(M n f) = isl(inl (F f)) BY (EquationFromHyp (-2) THEN Auto)) THEN Reduce -1) }
1
1. F : (ℕ ⟶ ℕ) ⟶ ℕ
2. M : n:ℕ ⟶ (ℕn ⟶ ℕ) ⟶ (ℕn?)
3. ∀f:ℕ ⟶ ℕ. (↓∃n:ℕ. (((M n f) = (inl (F f)) ∈ (ℕ?)) ∧ (∀m:ℕ. ((↑isl(M m f))
⇒ ((M m f) = (inl (F f)) ∈ (ℕ?))))))
4. f : ℕ ⟶ ℕ
5. n : ℕ
6. (M n f) = (inl (F f)) ∈ (ℕ?)
7. ∀m:ℕ. ((↑isl(M m f))
⇒ ((M m f) = (inl (F f)) ∈ (ℕ?)))
8. isl(M n f) = tt
⊢ <mu(λn.isl(M n f)), Ax> ∈ ∃n:ℕ. ((M n f) = (inl (F f)) ∈ (ℕ?))
Latex:
Latex:
1. F : (\mBbbN{} {}\mrightarrow{} \mBbbN{}) {}\mrightarrow{} \mBbbN{}
2. M : n:\mBbbN{} {}\mrightarrow{} (\mBbbN{}n {}\mrightarrow{} \mBbbN{}) {}\mrightarrow{} (\mBbbN{}n?)
3. \mforall{}f:\mBbbN{} {}\mrightarrow{} \mBbbN{}
(\mdownarrow{}\mexists{}n:\mBbbN{}. (((M n f) = (inl (F f))) \mwedge{} (\mforall{}m:\mBbbN{}. ((\muparrow{}isl(M m f)) {}\mRightarrow{} ((M m f) = (inl (F f)))))))
4. f : \mBbbN{} {}\mrightarrow{} \mBbbN{}
5. n : \mBbbN{}
6. (M n f) = (inl (F f))
7. \mforall{}m:\mBbbN{}. ((\muparrow{}isl(M m f)) {}\mRightarrow{} ((M m f) = (inl (F f))))
\mvdash{} <mu(\mlambda{}n.isl(M n f)), Ax> \mmember{} \mexists{}n:\mBbbN{}. ((M n f) = (inl (F f)))
By
Latex:
((Assert isl(M n f) = isl(inl (F f)) BY (EquationFromHyp (-2) THEN Auto)) THEN Reduce -1)
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