### Nuprl Lemma : stamps-example-ext

`∀n:ℕ. ∃i:ℕ. (∃j:ℕ [((n + 8) = ((3 * i) + (5 * j)) ∈ ℤ)])`

Proof

Definitions occuring in Statement :  nat: `ℕ` all: `∀x:A. B[x]` sq_exists: `∃x:A [B[x]]` exists: `∃x:A. B[x]` multiply: `n * m` add: `n + m` natural_number: `\$n` int: `ℤ` equal: `s = t ∈ T`
Definitions unfolded in proof :  member: `t ∈ T` stamps-example decidable__le decidable__and uall: `∀[x:A]. B[x]` so_lambda: `so_lambda(x,y,z,w.t[x; y; z; w])` so_apply: `x[s1;s2;s3;s4]` so_lambda: `λ2x.t[x]` top: `Top` so_apply: `x[s]` uimplies: `b supposing a` strict4: `strict4(F)` and: `P ∧ Q` all: `∀x:A. B[x]` implies: `P `` Q` has-value: `(a)↓` prop: `ℙ` guard: `{T}` or: `P ∨ Q` squash: `↓T` decidable__not decidable__implies decidable__less_than' decidable__false subtract: `n - m`
Lemmas referenced :  stamps-example lifting-strict-decide top_wf equal_wf has-value_wf_base base_wf is-exception_wf lifting-strict-less decidable__le decidable__and decidable__not decidable__implies decidable__less_than' decidable__false
Rules used in proof :  introduction sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity cut instantiate extract_by_obid hypothesis sqequalRule thin sqequalHypSubstitution isectElimination baseClosed isect_memberEquality voidElimination voidEquality independent_isectElimination independent_pairFormation lambdaFormation callbyvalueDecide hypothesisEquality equalityTransitivity equalitySymmetry unionEquality unionElimination sqleReflexivity dependent_functionElimination independent_functionElimination baseApply closedConclusion decideExceptionCases inrFormation because_Cache imageMemberEquality imageElimination exceptionSqequal inlFormation

Latex:
\mforall{}n:\mBbbN{}.  \mexists{}i:\mBbbN{}.  (\mexists{}j:\mBbbN{}  [((n  +  8)  =  ((3  *  i)  +  (5  *  j)))])

Date html generated: 2019_06_20-PM-01_05_04
Last ObjectModification: 2019_06_20-PM-01_00_03

Theory : int_1

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