Nuprl Lemma : double_sum_wf
∀[n,m:ℕ]. ∀[f:ℕn ⟶ ℕm ⟶ ℤ]. (sum(f[x;y] | x < n; y < m) ∈ ℤ)
Proof
Definitions occuring in Statement :
double_sum: sum(f[x; y] | x < n; y < m),
int_seg: {i..j-},
nat: ℕ,
uall: ∀[x:A]. B[x],
so_apply: x[s1;s2],
member: t ∈ T,
function: x:A ⟶ B[x],
natural_number: $n,
int: ℤ
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
double_sum: sum(f[x; y] | x < n; y < m),
so_lambda: λ2x.t[x],
so_apply: x[s1;s2],
nat: ℕ,
so_apply: x[s]
Lemmas referenced :
sum_wf,
int_seg_wf,
nat_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
introduction,
cut,
sqequalRule,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
lambdaEquality,
applyEquality,
natural_numberEquality,
setElimination,
rename,
hypothesis,
because_Cache,
axiomEquality,
equalityTransitivity,
equalitySymmetry,
Error :functionIsType,
Error :universeIsType,
intEquality,
isect_memberEquality,
functionEquality,
Error :inhabitedIsType
Latex:
\mforall{}[n,m:\mBbbN{}]. \mforall{}[f:\mBbbN{}n {}\mrightarrow{} \mBbbN{}m {}\mrightarrow{} \mBbbZ{}]. (sum(f[x;y] | x < n; y < m) \mmember{} \mBbbZ{})
Date html generated:
2019_06_20-PM-02_29_32
Last ObjectModification:
2018_09_26-PM-05_50_57
Theory : num_thy_1
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