Nuprl Lemma : p-reduce-0
∀p:ℕ+. ∀n:ℕ. (0 mod(p^n) = 0 ∈ ℤ)
Proof
Definitions occuring in Statement :
p-reduce: i mod(p^n),
nat_plus: ℕ+,
nat: ℕ,
all: ∀x:A. B[x],
natural_number: $n,
int: ℤ,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x],
p-reduce: i mod(p^n),
uall: ∀[x:A]. B[x],
member: t ∈ T,
int_seg: {i..j-},
lelt: i ≤ j < k,
and: P ∧ Q,
le: A ≤ B,
less_than': less_than'(a;b),
false: False,
not: ¬A,
implies: P ⇒ Q,
prop: ℙ,
nat_plus: ℕ+
Lemmas referenced :
modulus_base,
exp_wf_nat_plus,
false_wf,
exp-positive-stronger,
lelt_wf,
exp_wf2,
nat_wf,
nat_plus_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaFormation,
cut,
sqequalRule,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
hypothesis,
dependent_set_memberEquality,
natural_numberEquality,
independent_pairFormation,
dependent_functionElimination,
setElimination,
rename,
because_Cache
Latex:
\mforall{}p:\mBbbN{}\msupplus{}. \mforall{}n:\mBbbN{}. (0 mod(p\^{}n) = 0)
Date html generated:
2018_05_21-PM-03_18_00
Last ObjectModification:
2018_05_19-AM-08_08_58
Theory : rings_1
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