Nuprl Lemma : div_nat_induction-ext
∀b:{b:ℤ| 1 < b} . ∀[P:ℕ ⟶ ℙ]. (P[0] ⇒ (∀i:ℕ+. (P[i ÷ b] ⇒ P[i])) ⇒ (∀i:ℕ. P[i]))
Proof
Definitions occuring in Statement :
nat_plus: ℕ+,
nat: ℕ,
less_than: a < b,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_apply: x[s],
all: ∀x:A. B[x],
implies: P ⇒ Q,
set: {x:A| B[x]} ,
function: x:A ⟶ B[x],
divide: n ÷ m,
natural_number: $n,
int: ℤ
Definitions unfolded in proof :
member: t ∈ T,
so_apply: x[s1;s2],
div_nat_induction,
decidable__equal_int,
decidable__int_equal,
uall: ∀[x:A]. B[x],
so_lambda: so_lambda(x,y,z,w.t[x; y; z; w]),
so_apply: x[s1;s2;s3;s4],
so_lambda: λ2x.t[x],
top: Top,
so_apply: x[s],
uimplies: b supposing a
Lemmas referenced :
div_nat_induction,
lifting-strict-int_eq,
istype-void,
strict4-decide,
decidable__equal_int,
decidable__int_equal
Rules used in proof :
introduction,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
cut,
instantiate,
extract_by_obid,
hypothesis,
sqequalRule,
thin,
sqequalHypSubstitution,
equalityTransitivity,
equalitySymmetry,
isectElimination,
baseClosed,
Error :isect_memberEquality_alt,
voidElimination,
independent_isectElimination
Latex:
\mforall{}b:\{b:\mBbbZ{}| 1 < b\} . \mforall{}[P:\mBbbN{} {}\mrightarrow{} \mBbbP{}]. (P[0] {}\mRightarrow{} (\mforall{}i:\mBbbN{}\msupplus{}. (P[i \mdiv{} b] {}\mRightarrow{} P[i])) {}\mRightarrow{} (\mforall{}i:\mBbbN{}. P[i]))
Date html generated:
2019_06_20-PM-02_33_18
Last ObjectModification:
2019_04_15-PM-10_31_58
Theory : num_thy_1
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