Nuprl Lemma : bag-combine-eq-right
∀[A,B:Type]. ∀[b:bag(A)]. ∀[f1,f2:A ⟶ bag(B)].
⋃x∈b.f1[x] = ⋃x∈b.f2[x] ∈ bag(B) supposing ∀x:{x:A| x ↓∈ b} . (f1[x] = f2[x] ∈ bag(B))
Proof
Definitions occuring in Statement :
bag-member: x ↓∈ bs,
bag-combine: ⋃x∈bs.f[x],
bag: bag(T),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
all: ∀x:A. B[x],
set: {x:A| B[x]} ,
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
bag-combine: ⋃x∈bs.f[x],
uall: ∀[x:A]. B[x],
member: t ∈ T,
squash: ↓T,
exists: ∃x:A. B[x],
prop: ℙ,
so_lambda: λ2x.t[x],
so_apply: x[s],
bag-map: bag-map(f;bs),
true: True,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
guard: {T},
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q,
implies: P ⇒ Q,
all: ∀x:A. B[x],
nat: ℕ,
int_seg: {i..j-},
lelt: i ≤ j < k,
le: A ≤ B
Lemmas referenced :
bag_to_squash_list,
all_wf,
bag-member_wf,
equal_wf,
bag_wf,
bag-map_wf,
bag-union_wf,
squash_wf,
true_wf,
iff_weakening_equal,
map_equal,
select_wf,
bag-member-select,
lelt_wf,
length_wf,
list-subtype-bag,
less_than_wf,
nat_wf,
length_wf_nat,
map_wf,
subtype_rel_self,
subtype_rel_set,
list_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
because_Cache,
hypothesisEquality,
imageElimination,
productElimination,
promote_hyp,
hypothesis,
equalitySymmetry,
hyp_replacement,
applyLambdaEquality,
setEquality,
cumulativity,
sqequalRule,
lambdaEquality,
applyEquality,
functionExtensionality,
setElimination,
rename,
natural_numberEquality,
equalityTransitivity,
functionEquality,
universeEquality,
isect_memberFormation,
isect_memberEquality,
axiomEquality,
imageMemberEquality,
baseClosed,
independent_isectElimination,
independent_functionElimination,
lambdaFormation,
dependent_functionElimination,
dependent_set_memberEquality,
independent_pairFormation
Latex:
\mforall{}[A,B:Type]. \mforall{}[b:bag(A)]. \mforall{}[f1,f2:A {}\mrightarrow{} bag(B)].
\mcup{}x\mmember{}b.f1[x] = \mcup{}x\mmember{}b.f2[x] supposing \mforall{}x:\{x:A| x \mdownarrow{}\mmember{} b\} . (f1[x] = f2[x])
Date html generated:
2017_10_01-AM-08_56_09
Last ObjectModification:
2017_07_26-PM-04_38_11
Theory : bags
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