Nuprl Lemma : p-sep-irrefl

[p:ℕ+]. ∀[x:p-adics(p)].  p-sep(x;x))


Proof




Definitions occuring in Statement :  p-sep: p-sep(x;y) p-adics: p-adics(p) nat_plus: + uall: [x:A]. B[x] not: ¬A
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T not: ¬A implies:  Q false: False p-sep: p-sep(x;y) exists: x:A. B[x] prop: nat_plus: + p-adics: p-adics(p) subtype_rel: A ⊆B int_seg: {i..j-}
Lemmas referenced :  p-sep_wf p-adics_wf nat_plus_wf int_seg_wf exp_wf2 nat_plus_subtype_nat
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut lambdaFormation thin sqequalHypSubstitution productElimination independent_functionElimination voidElimination because_Cache hypothesis extract_by_obid isectElimination hypothesisEquality sqequalRule lambdaEquality dependent_functionElimination setElimination rename isect_memberEquality applyEquality natural_numberEquality

Latex:
\mforall{}[p:\mBbbN{}\msupplus{}].  \mforall{}[x:p-adics(p)].    (\mneg{}p-sep(x;x))



Date html generated: 2018_05_21-PM-03_23_18
Last ObjectModification: 2018_05_19-AM-08_21_35

Theory : rings_1


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