Nuprl Lemma : bag-remove1-property
∀[T:Type]
∀eq:EqDecider(T). ∀x:T. ∀bs:bag(T).
((∃as:bag(T). ((bs = ({x} + as) ∈ bag(T)) ∧ (bag-remove1(eq;bs;x) = (inl as) ∈ (bag(T)?))))
∨ ((¬x ↓∈ bs) ∧ (bag-remove1(eq;bs;x) = (inr ⋅ ) ∈ (bag(T)?))))
Proof
Definitions occuring in Statement :
bag-remove1: bag-remove1(eq;bs;a),
bag-member: x ↓∈ bs,
bag-append: as + bs,
single-bag: {x},
bag: bag(T),
deq: EqDecider(T),
it: ⋅,
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
not: ¬A,
or: P ∨ Q,
and: P ∧ Q,
unit: Unit,
inr: inr x ,
inl: inl x,
union: left + right,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
member: t ∈ T,
implies: P ⇒ Q,
decidable: Dec(P),
or: P ∨ Q,
and: P ∧ Q,
not: ¬A,
false: False,
squash: ↓T,
exists: ∃x:A. B[x],
prop: ℙ,
isl: isl(x),
outl: outl(x),
uimplies: b supposing a,
subtype_rel: A ⊆r B,
assert: ↑b,
ifthenelse: if b then t else f fi ,
btrue: tt,
cand: A c∧ B,
true: True,
iff: P ⇐⇒ Q,
bag-append: as + bs,
nat: ℕ,
so_lambda: λ2x.t[x],
so_apply: x[s],
guard: {T},
rev_implies: P ⇐ Q,
single-bag: {x},
top: Top,
sq_stable: SqStable(P),
bfalse: ff,
bag-member: x ↓∈ bs,
respects-equality: respects-equality(S;T)
Lemmas referenced :
decidable__bag-member,
decidable-equal-deq,
bag_wf,
deq_wf,
istype-universe,
istype-void,
bag_to_squash_list,
bag-member_wf,
decidable_wf,
squash_wf,
assert_wf,
bag-remove1_wf,
btrue_wf,
bfalse_wf,
equal_wf,
bag-append_wf,
single-bag_wf,
assert_elim,
unit_subtype_base,
btrue_neq_bfalse,
bag-remove1-property1,
isl_wf,
list_wf,
unit_wf2,
list-subtype-bag,
outl_wf,
bag-member-list,
length_wf_nat,
istype-nat,
set_subtype_base,
le_wf,
istype-int,
int_subtype_base,
true_wf,
bag-append-comm,
cons_wf,
nil_wf,
reverse_wf,
subtype_rel_self,
iff_weakening_equal,
bag-append-assoc2,
reverse-bag,
subtype_rel_set,
equal-wf-base,
bag_qinc,
sq_stable__and,
sq_stable__assert,
sq_stable__equal,
istype-assert,
not_wf,
equal-wf-T-base,
subtype_rel_union,
equal_functionality_wrt_subtype_rel2,
member_append,
append_wf,
cons_member,
l_member_wf,
subtype-respects-equality
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation_alt,
lambdaFormation_alt,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
independent_functionElimination,
because_Cache,
hypothesis,
dependent_functionElimination,
universeIsType,
instantiate,
universeEquality,
unionElimination,
inlFormation_alt,
sqequalRule,
productIsType,
functionIsType,
inhabitedIsType,
voidElimination,
imageElimination,
productElimination,
promote_hyp,
equalitySymmetry,
hyp_replacement,
applyLambdaEquality,
functionEquality,
rename,
imageMemberEquality,
baseClosed,
productEquality,
equalityIstype,
equalityTransitivity,
independent_isectElimination,
dependent_set_memberEquality_alt,
independent_pairFormation,
baseApply,
closedConclusion,
applyEquality,
sqequalBase,
setElimination,
natural_numberEquality,
lambdaEquality_alt,
intEquality,
isect_memberEquality_alt,
axiomEquality,
functionIsTypeImplies,
dependent_pairFormation_alt,
inlEquality_alt,
inrFormation_alt,
unionIsType,
unionEquality
Latex:
\mforall{}[T:Type]
\mforall{}eq:EqDecider(T). \mforall{}x:T. \mforall{}bs:bag(T).
((\mexists{}as:bag(T). ((bs = (\{x\} + as)) \mwedge{} (bag-remove1(eq;bs;x) = (inl as))))
\mvee{} ((\mneg{}x \mdownarrow{}\mmember{} bs) \mwedge{} (bag-remove1(eq;bs;x) = (inr \mcdot{} ))))
Date html generated:
2019_10_16-AM-11_30_46
Last ObjectModification:
2018_11_30-PM-00_20_42
Theory : bags_2
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