Nuprl Lemma : complete-nat-induction

[P:ℕ ⟶ ℙ]. ((∀n:ℕ((∀m:ℕn. P[m])  P[n]))  (∀n:ℕP[n]))


Proof




Definitions occuring in Statement :  int_seg: {i..j-} nat: uall: [x:A]. B[x] prop: so_apply: x[s] all: x:A. B[x] implies:  Q function: x:A ⟶ B[x] natural_number: $n
Definitions unfolded in proof :  uall: [x:A]. B[x] implies:  Q all: x:A. B[x] member: t ∈ T prop: so_lambda: λ2x.t[x] nat: so_apply: x[s] subtype_rel: A ⊆B uimplies: supposing a le: A ≤ B and: P ∧ Q less_than': less_than'(a;b) false: False not: ¬A so_lambda: λ2y.t[x; y] so_apply: x[s1;s2]
Lemmas referenced :  all_wf nat_wf int_seg_wf int_seg_subtype_nat false_wf natrec_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation lambdaFormation sqequalHypSubstitution sqequalRule rename cut lemma_by_obid isectElimination thin hypothesis lambdaEquality functionEquality natural_numberEquality setElimination hypothesisEquality applyEquality independent_isectElimination independent_pairFormation because_Cache universeEquality cumulativity introduction

Latex:
\mforall{}[P:\mBbbN{}  {}\mrightarrow{}  \mBbbP{}].  ((\mforall{}n:\mBbbN{}.  ((\mforall{}m:\mBbbN{}n.  P[m])  {}\mRightarrow{}  P[n]))  {}\mRightarrow{}  (\mforall{}n:\mBbbN{}.  P[n]))



Date html generated: 2016_05_13-PM-04_03_17
Last ObjectModification: 2015_12_26-AM-10_56_08

Theory : int_1


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