Nuprl Lemma : bag-sum-nonneg

[A:Type]. ∀[f:A ⟶ ℤ]. ∀[ba:bag(A)].  0 ≤ bag-sum(ba;x.f[x]) supposing ∀x:A. (0 ≤ f[x])


Proof




Definitions occuring in Statement :  bag-sum: bag-sum(ba;x.f[x]) bag: bag(T) uimplies: supposing a uall: [x:A]. B[x] so_apply: x[s] le: A ≤ B all: x:A. B[x] function: x:A ⟶ B[x] natural_number: $n int: universe: Type
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a so_lambda: λ2x.t[x] nat: so_apply: x[s] prop: le: A ≤ B and: P ∧ Q not: ¬A implies:  Q false: False guard: {T} all: x:A. B[x]
Lemmas referenced :  zero-le-nat bag-sum_wf_nat le_wf less_than'_wf bag-sum_wf all_wf bag_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut lemma_by_obid sqequalHypSubstitution isectElimination thin cumulativity hypothesisEquality sqequalRule lambdaEquality dependent_set_memberEquality applyEquality natural_numberEquality hypothesis because_Cache productElimination independent_pairEquality dependent_functionElimination axiomEquality equalityTransitivity equalitySymmetry isect_memberEquality functionEquality intEquality voidElimination universeEquality

Latex:
\mforall{}[A:Type].  \mforall{}[f:A  {}\mrightarrow{}  \mBbbZ{}].  \mforall{}[ba:bag(A)].    0  \mleq{}  bag-sum(ba;x.f[x])  supposing  \mforall{}x:A.  (0  \mleq{}  f[x])



Date html generated: 2016_05_15-PM-02_29_35
Last ObjectModification: 2015_12_27-AM-09_49_30

Theory : bags


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