Nuprl Lemma : p-sep_wf

[p:ℕ+]. ∀[x,y:p-adics(p)].  (p-sep(x;y) ∈ ℙ)


Proof




Definitions occuring in Statement :  p-sep: p-sep(x;y) p-adics: p-adics(p) nat_plus: + uall: [x:A]. B[x] prop: member: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T p-sep: p-sep(x;y) so_lambda: λ2x.t[x] p-adics: p-adics(p) subtype_rel: A ⊆B int_seg: {i..j-} nat_plus: + so_apply: x[s]
Lemmas referenced :  exists_wf nat_plus_wf not_wf equal_wf int_seg_wf exp_wf2 nat_plus_subtype_nat p-adics_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule extract_by_obid sqequalHypSubstitution isectElimination thin hypothesis lambdaEquality intEquality applyEquality setElimination rename hypothesisEquality natural_numberEquality because_Cache axiomEquality equalityTransitivity equalitySymmetry isect_memberEquality

Latex:
\mforall{}[p:\mBbbN{}\msupplus{}].  \mforall{}[x,y:p-adics(p)].    (p-sep(x;y)  \mmember{}  \mBbbP{})



Date html generated: 2018_05_21-PM-03_23_09
Last ObjectModification: 2018_05_19-AM-08_21_25

Theory : rings_1


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