Nuprl Lemma : mkpadic_wf

[p:ℕ+]. ∀[n:ℕ]. ∀[a:p-adics(p)].  ((a/p^n) ∈ padic(p))


Proof




Definitions occuring in Statement :  mkpadic: (a/p^n) padic: padic(p) p-adics: p-adics(p) nat_plus: + nat: uall: [x:A]. B[x] member: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T mkpadic: (a/p^n) basic-padic: basic-padic(p) nat_plus: +
Lemmas referenced :  bpa-norm_wf_padic p-adics_wf nat_wf nat_plus_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut sqequalRule extract_by_obid sqequalHypSubstitution isectElimination thin hypothesisEquality independent_pairEquality hypothesis axiomEquality equalityTransitivity equalitySymmetry setElimination rename isect_memberEquality because_Cache

Latex:
\mforall{}[p:\mBbbN{}\msupplus{}].  \mforall{}[n:\mBbbN{}].  \mforall{}[a:p-adics(p)].    ((a/p\^{}n)  \mmember{}  padic(p))



Date html generated: 2018_05_21-PM-03_26_28
Last ObjectModification: 2018_05_19-AM-08_23_34

Theory : rings_1


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