Nuprl Lemma : regularize-regular

k:ℕ+. ∀x:{f:ℕ+ ⟶ ℤk-regular-seq(f)} .  (regularize(k;x) x ∈ (ℕ+ ⟶ ℤ))


Proof




Definitions occuring in Statement :  regularize: regularize(k;f) regular-int-seq: k-regular-seq(f) nat_plus: + all: x:A. B[x] set: {x:A| B[x]}  function: x:A ⟶ B[x] int: equal: t ∈ T
Definitions unfolded in proof :  all: x:A. B[x] member: t ∈ T subtype_rel: A ⊆B implies:  Q regular-int-seq: k-regular-seq(f) uall: [x:A]. B[x] nat: nat_plus: + decidable: Dec(P) or: P ∨ Q uimplies: supposing a not: ¬A satisfiable_int_formula: satisfiable_int_formula(fmla) exists: x:A. B[x] false: False and: P ∧ Q prop: iff: ⇐⇒ Q rev_implies:  Q guard: {T} int_seg: {i..j-} lelt: i ≤ j < k le: A ≤ B less_than: a < b squash: T subtract: m

Latex:
\mforall{}k:\mBbbN{}\msupplus{}.  \mforall{}x:\{f:\mBbbN{}\msupplus{}  {}\mrightarrow{}  \mBbbZ{}|  k-regular-seq(f)\}  .    (regularize(k;x)  =  x)



Date html generated: 2020_05_20-AM-11_17_44
Last ObjectModification: 2020_03_14-AM-09_31_36

Theory : reals


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