Step
*
1
1
of Lemma
strong-fun-connected-induction
.....assertion.....
1. [T] : Type
2. f : T ⟶ T
3. [R] : T ⟶ T ⟶ ℙ
4. h : T ⟶ ℕ
5. ∀x:T. (((f x) = x ∈ T) ∨ h (f x) < h x)
6. ∀x:T. R[x;x]
7. ∀x,y,z:T.
(y is f*(z)
⇒ (∀u:T. (y is f*(u)
⇒ u is f*(z)
⇒ R[u;z]))
⇒ R[x;z]) supposing
((¬(x = y ∈ T)) and
(x = (f y) ∈ T))
⊢ ∀n:ℕ. ∀x,y:T. (x is f*(y)
⇒ (h y) - h x < n
⇒ R[x;y])
BY
{ InductionOnNat }
1
.....basecase.....
1. [T] : Type
2. f : T ⟶ T
3. [R] : T ⟶ T ⟶ ℙ
4. h : T ⟶ ℕ
5. ∀x:T. (((f x) = x ∈ T) ∨ h (f x) < h x)
6. ∀x:T. R[x;x]
7. ∀x,y,z:T.
(y is f*(z)
⇒ (∀u:T. (y is f*(u)
⇒ u is f*(z)
⇒ R[u;z]))
⇒ R[x;z]) supposing
((¬(x = y ∈ T)) and
(x = (f y) ∈ T))
⊢ ∀x,y:T. (x is f*(y)
⇒ (h y) - h x < 0
⇒ R[x;y])
2
.....upcase.....
1. [T] : Type
2. f : T ⟶ T
3. [R] : T ⟶ T ⟶ ℙ
4. h : T ⟶ ℕ
5. ∀x:T. (((f x) = x ∈ T) ∨ h (f x) < h x)
6. ∀x:T. R[x;x]
7. ∀x,y,z:T.
(y is f*(z)
⇒ (∀u:T. (y is f*(u)
⇒ u is f*(z)
⇒ R[u;z]))
⇒ R[x;z]) supposing
((¬(x = y ∈ T)) and
(x = (f y) ∈ T))
8. n : ℤ
9. [%4] : 0 < n
10. ∀x,y:T. (x is f*(y)
⇒ (h y) - h x < n - 1
⇒ R[x;y])
⊢ ∀x,y:T. (x is f*(y)
⇒ (h y) - h x < n
⇒ R[x;y])
Latex:
Latex:
.....assertion.....
1. [T] : Type
2. f : T {}\mrightarrow{} T
3. [R] : T {}\mrightarrow{} T {}\mrightarrow{} \mBbbP{}
4. h : T {}\mrightarrow{} \mBbbN{}
5. \mforall{}x:T. (((f x) = x) \mvee{} h (f x) < h x)
6. \mforall{}x:T. R[x;x]
7. \mforall{}x,y,z:T.
(y is f*(z) {}\mRightarrow{} (\mforall{}u:T. (y is f*(u) {}\mRightarrow{} u is f*(z) {}\mRightarrow{} R[u;z])) {}\mRightarrow{} R[x;z]) supposing
((\mneg{}(x = y)) and
(x = (f y)))
\mvdash{} \mforall{}n:\mBbbN{}. \mforall{}x,y:T. (x is f*(y) {}\mRightarrow{} (h y) - h x < n {}\mRightarrow{} R[x;y])
By
Latex:
InductionOnNat
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