Nuprl Lemma : member-int_seg
∀[i,j,x:ℤ]. (x ∈ {i..j-}) supposing ((i ≤ x) and x < j)
Proof
Definitions occuring in Statement :
int_seg: {i..j-},
less_than: a < b,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
le: A ≤ B,
member: t ∈ T,
int: ℤ
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
int_seg: {i..j-},
lelt: i ≤ j < k,
and: P ∧ Q
Lemmas referenced :
istype-le,
istype-less_than,
istype-int
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
introduction,
cut,
Error :dependent_set_memberEquality_alt,
hypothesisEquality,
independent_pairFormation,
hypothesis,
sqequalRule,
Error :productIsType,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
because_Cache,
axiomEquality,
equalityTransitivity,
equalitySymmetry,
Error :isect_memberEquality_alt,
Error :isectIsTypeImplies,
Error :inhabitedIsType
Latex:
\mforall{}[i,j,x:\mBbbZ{}]. (x \mmember{} \{i..j\msupminus{}\}) supposing ((i \mleq{} x) and x < j)
Date html generated:
2019_06_20-AM-11_23_46
Last ObjectModification:
2018_10_25-PM-00_59_43
Theory : arithmetic
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