Step
*
1
1
of Lemma
closures-meet
.....assertion.....
1. [P] : ℝ ⟶ ℙ
2. [Q] : ℝ ⟶ ℙ
3. a0 : ℝ
4. b0 : ℝ
5. P a0
6. Q b0
7. a0 ≤ b0
8. c : ℝ
9. r0 ≤ c
10. c < r1
11. ∀a,b:ℝ.
(((P a) ∧ (Q b) ∧ (a ≤ b))
⇒ (∃a',b':ℝ. ((P a') ∧ (Q b') ∧ (a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c)))))
⊢ ∃a,b:ℕ ⟶ ℝ
∀n:ℕ
((P a[n])
∧ (Q b[n])
∧ (a[n] ≤ a[n + 1])
∧ (a[n + 1] ≤ b[n + 1])
∧ (b[n + 1] ≤ b[n])
∧ ((b[n + 1] - a[n + 1]) ≤ ((b[n] - a[n]) * c)))
BY
{ Assert ⌜∀abp:a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
∃abp':a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
let a,b,p = abp in
let a',b',p' = abp' in
(a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c))⌝⋅ }
1
.....assertion.....
1. [P] : ℝ ⟶ ℙ
2. [Q] : ℝ ⟶ ℙ
3. a0 : ℝ
4. b0 : ℝ
5. P a0
6. Q b0
7. a0 ≤ b0
8. c : ℝ
9. r0 ≤ c
10. c < r1
11. ∀a,b:ℝ.
(((P a) ∧ (Q b) ∧ (a ≤ b))
⇒ (∃a',b':ℝ. ((P a') ∧ (Q b') ∧ (a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c)))))
⊢ ∀abp:a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
∃abp':a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
let a,b,p = abp in
let a',b',p' = abp' in
(a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c))
2
1. [P] : ℝ ⟶ ℙ
2. [Q] : ℝ ⟶ ℙ
3. a0 : ℝ
4. b0 : ℝ
5. P a0
6. Q b0
7. a0 ≤ b0
8. c : ℝ
9. r0 ≤ c
10. c < r1
11. ∀a,b:ℝ.
(((P a) ∧ (Q b) ∧ (a ≤ b))
⇒ (∃a',b':ℝ. ((P a') ∧ (Q b') ∧ (a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c)))))
12. ∀abp:a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
∃abp':a:ℝ × b:ℝ × ((P a) ∧ (Q b) ∧ (a ≤ b))
let a,b,p = abp in
let a',b',p' = abp' in
(a ≤ a') ∧ (a' ≤ b') ∧ (b' ≤ b) ∧ ((b' - a') ≤ ((b - a) * c))
⊢ ∃a,b:ℕ ⟶ ℝ
∀n:ℕ
((P a[n])
∧ (Q b[n])
∧ (a[n] ≤ a[n + 1])
∧ (a[n + 1] ≤ b[n + 1])
∧ (b[n + 1] ≤ b[n])
∧ ((b[n + 1] - a[n + 1]) ≤ ((b[n] - a[n]) * c)))
Latex:
Latex:
.....assertion.....
1. [P] : \mBbbR{} {}\mrightarrow{} \mBbbP{}
2. [Q] : \mBbbR{} {}\mrightarrow{} \mBbbP{}
3. a0 : \mBbbR{}
4. b0 : \mBbbR{}
5. P a0
6. Q b0
7. a0 \mleq{} b0
8. c : \mBbbR{}
9. r0 \mleq{} c
10. c < r1
11. \mforall{}a,b:\mBbbR{}.
(((P a) \mwedge{} (Q b) \mwedge{} (a \mleq{} b))
{}\mRightarrow{} (\mexists{}a',b':\mBbbR{}
((P a') \mwedge{} (Q b') \mwedge{} (a \mleq{} a') \mwedge{} (a' \mleq{} b') \mwedge{} (b' \mleq{} b) \mwedge{} ((b' - a') \mleq{} ((b - a) * c)))))
\mvdash{} \mexists{}a,b:\mBbbN{} {}\mrightarrow{} \mBbbR{}
\mforall{}n:\mBbbN{}
((P a[n])
\mwedge{} (Q b[n])
\mwedge{} (a[n] \mleq{} a[n + 1])
\mwedge{} (a[n + 1] \mleq{} b[n + 1])
\mwedge{} (b[n + 1] \mleq{} b[n])
\mwedge{} ((b[n + 1] - a[n + 1]) \mleq{} ((b[n] - a[n]) * c)))
By
Latex:
Assert \mkleeneopen{}\mforall{}abp:a:\mBbbR{} \mtimes{} b:\mBbbR{} \mtimes{} ((P a) \mwedge{} (Q b) \mwedge{} (a \mleq{} b))
\mexists{}abp':a:\mBbbR{} \mtimes{} b:\mBbbR{} \mtimes{} ((P a) \mwedge{} (Q b) \mwedge{} (a \mleq{} b))
let a,b,p = abp in
let a',b',p' = abp' in
(a \mleq{} a') \mwedge{} (a' \mleq{} b') \mwedge{} (b' \mleq{} b) \mwedge{} ((b' - a') \mleq{} ((b - a) * c))\mkleeneclose{}\mcdot{}
Home
Index