Nuprl Lemma : regularize-k-regular
∀k:ℕ+. ∀f:ℕ+ ⟶ ℤ. k-regular-seq(regularize(k;f))
Proof
Definitions occuring in Statement :
regularize: regularize(k;f),
regular-int-seq: k-regular-seq(f),
nat_plus: ℕ+,
all: ∀x:A. B[x],
function: x:A ⟶ B[x],
int: ℤ
Definitions unfolded in proof :
absval: |i|,
so_apply: x[s],
so_lambda: λ2x.t[x],
squash: ↓T,
less_than: a < b,
lelt: i ≤ j < k,
int_seg: {i..j-},
less_than': less_than'(a;b),
le: A ≤ B,
subtype_rel: A ⊆r B,
ge: i ≥ j ,
assert: ↑b,
bnot: ¬bb,
guard: {T},
sq_type: SQType(T),
bfalse: ff,
ifthenelse: if b then t else f fi ,
uiff: uiff(P;Q),
btrue: tt,
it: ⋅,
unit: Unit,
bool: 𝔹,
prop: ℙ,
false: False,
exists: ∃x:A. B[x],
satisfiable_int_formula: satisfiable_int_formula(fmla),
not: ¬A,
uimplies: b supposing a,
or: P ∨ Q,
decidable: Dec(P),
nat_plus: ℕ+,
nat: ℕ,
regularize: regularize(k;f),
implies: P ⇒ Q,
rev_implies: P ⇐ Q,
and: P ∧ Q,
iff: P ⇐⇒ Q,
uall: ∀[x:A]. B[x],
member: t ∈ T,
all: ∀x:A. B[x],
subtract: n - m,
has-value: (a)↓,
true: True,
top: Top,
cand: A c∧ B,
let: let,
rneq: x ≠ y,
nequal: a ≠ b ∈ T ,
int_nzero: ℤ-o,
rational-approx: (x within 1/n),
int-to-real: r(n),
int-rdiv: (a)/k1,
rev_uimplies: rev_uimplies(P;Q)
Latex:
\mforall{}k:\mBbbN{}\msupplus{}. \mforall{}f:\mBbbN{}\msupplus{} {}\mrightarrow{} \mBbbZ{}. k-regular-seq(regularize(k;f))
Date html generated:
2020_05_20-AM-11_17_29
Last ObjectModification:
2019_12_28-AM-11_34_51
Theory : reals
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