Nuprl Lemma : absval_nat_plus
∀[x:ℤ]. |x| ∈ ℕ+ supposing ¬(x = 0 ∈ ℤ)
Proof
Definitions occuring in Statement : 
absval: |i|
, 
nat_plus: ℕ+
, 
uimplies: b supposing a
, 
uall: ∀[x:A]. B[x]
, 
not: ¬A
, 
member: t ∈ T
, 
natural_number: $n
, 
int: ℤ
, 
equal: s = t ∈ T
Definitions unfolded in proof : 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
uimplies: b supposing a
, 
nat_plus: ℕ+
, 
subtype_rel: A ⊆r B
, 
nat: ℕ
, 
prop: ℙ
, 
uiff: uiff(P;Q)
, 
and: P ∧ Q
, 
rev_uimplies: rev_uimplies(P;Q)
, 
nequal: a ≠ b ∈ T 
Lemmas referenced : 
absval_wf, 
nat_wf, 
less_than_wf, 
not_wf, 
equal_wf, 
absval-positive
Rules used in proof : 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
sqequalReflexivity, 
isect_memberFormation, 
introduction, 
cut, 
dependent_set_memberEquality, 
lemma_by_obid, 
sqequalHypSubstitution, 
isectElimination, 
thin, 
hypothesisEquality, 
hypothesis, 
applyEquality, 
lambdaEquality, 
setElimination, 
rename, 
sqequalRule, 
natural_numberEquality, 
axiomEquality, 
equalityTransitivity, 
equalitySymmetry, 
intEquality, 
isect_memberEquality, 
because_Cache, 
productElimination, 
independent_isectElimination
Latex:
\mforall{}[x:\mBbbZ{}].  |x|  \mmember{}  \mBbbN{}\msupplus{}  supposing  \mneg{}(x  =  0)
Date html generated:
2016_05_14-AM-07_20_37
Last ObjectModification:
2015_12_26-PM-01_32_12
Theory : int_2
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