Nuprl Lemma : real-continuity1

a,b:ℝ.
  ∀f:[a, b] ⟶ℝ
    ∀k:ℕ+. ∃d:{d:ℝr0 < d} . ∀x,y:{x:ℝx ∈ [a, b]} .  ((|x y| ≤ d)  (|(f x) y| ≤ (r1/r(k)))) 
    supposing ∀x,y:{x:ℝx ∈ [a, b]} .  ((x y)  ((f x) (f y))) 
  supposing a ≤ b


Proof




Definitions occuring in Statement :  rfun: I ⟶ℝ rccint: [l, u] i-member: r ∈ I rdiv: (x/y) rleq: x ≤ y rless: x < y rabs: |x| rsub: y req: y int-to-real: r(n) real: nat_plus: + uimplies: supposing a all: x:A. B[x] exists: x:A. B[x] implies:  Q set: {x:A| B[x]}  apply: a natural_number: $n
Definitions unfolded in proof :  real-fun: real-fun(f;a;b) real-cont: real-cont(f;a;b)
Lemmas referenced :  real-continuity-ext
Rules used in proof :  hypothesis extract_by_obid introduction cut computationStep sqequalTransitivity sqequalReflexivity sqequalRule sqequalSubstitution

Latex:
\mforall{}a,b:\mBbbR{}.
    \mforall{}f:[a,  b]  {}\mrightarrow{}\mBbbR{}
        \mforall{}k:\mBbbN{}\msupplus{}
            \mexists{}d:\{d:\mBbbR{}|  r0  <  d\}  .  \mforall{}x,y:\{x:\mBbbR{}|  x  \mmember{}  [a,  b]\}  .    ((|x  -  y|  \mleq{}  d)  {}\mRightarrow{}  (|(f  x)  -  f  y|  \mleq{}  (r1/r(k)))) 
        supposing  \mforall{}x,y:\{x:\mBbbR{}|  x  \mmember{}  [a,  b]\}  .    ((x  =  y)  {}\mRightarrow{}  ((f  x)  =  (f  y))) 
    supposing  a  \mleq{}  b



Date html generated: 2018_05_22-PM-02_11_54
Last ObjectModification: 2018_05_21-AM-00_27_53

Theory : reals


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