Nuprl Lemma : real-continuity1-ext
∀a,b:ℝ.
  ∀f:[a, b] ⟶ℝ
    ∀k:ℕ+. ∃d:{d:ℝ| r0 < d} . ∀x,y:{x:ℝ| x ∈ [a, b]} .  ((|x - y| ≤ d) 
⇒ (|(f x) - f y| ≤ (r1/r(k)))) 
    supposing ∀x,y:{x:ℝ| x ∈ [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: x - y
, 
req: x = y
, 
int-to-real: r(n)
, 
real: ℝ
, 
nat_plus: ℕ+
, 
uimplies: b supposing a
, 
all: ∀x:A. B[x]
, 
exists: ∃x:A. B[x]
, 
implies: P 
⇒ Q
, 
set: {x:A| B[x]} 
, 
apply: f a
, 
natural_number: $n
Definitions unfolded in proof : 
real-continuity-ext, 
real-continuity1, 
so_apply: x[s]
, 
so_lambda: λ2x.t[x]
, 
guard: {T}
, 
implies: P 
⇒ Q
, 
all: ∀x:A. B[x]
, 
sq_type: SQType(T)
, 
unit: Unit
, 
bool: 𝔹
, 
uimplies: b supposing a
, 
uall: ∀[x:A]. B[x]
, 
ifthenelse: if b then t else f fi 
, 
bfalse: ff
, 
it: ⋅
, 
btrue: tt
, 
subtract: n - m
, 
member: t ∈ T
Lemmas referenced : 
product_subtype_base, 
unit_subtype_base, 
unit_wf2, 
trivial-equal, 
bool_subtype_base, 
bool_wf, 
subtype_base_sq, 
real-continuity1, 
real-continuity-ext
Rules used in proof : 
independent_pairEquality, 
lambdaFormation, 
lambdaEquality, 
productEquality, 
inrEquality, 
because_Cache, 
independent_functionElimination, 
dependent_functionElimination, 
natural_numberEquality, 
axiomEquality, 
inlEquality, 
independent_isectElimination, 
cumulativity, 
isectElimination, 
equalitySymmetry, 
equalityTransitivity, 
sqequalHypSubstitution, 
thin, 
sqequalRule, 
hypothesis, 
extract_by_obid, 
instantiate, 
cut, 
sqequalReflexivity, 
computationStep, 
sqequalTransitivity, 
sqequalSubstitution, 
introduction
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_12_07
Last ObjectModification:
2018_05_21-AM-00_29_07
Theory : reals
Home
Index