Nuprl Lemma : bag-subtract-append
∀[T:Type]. ∀[eq:EqDecider(T)]. ∀[as,bs:bag(T)].  (bag-subtract(eq;as + bs;as) = bs ∈ bag(T))
Proof
Definitions occuring in Statement : 
bag-subtract: bag-subtract(eq;bs;as)
, 
bag-append: as + bs
, 
bag: bag(T)
, 
deq: EqDecider(T)
, 
uall: ∀[x:A]. B[x]
, 
universe: Type
, 
equal: s = t ∈ T
Definitions unfolded in proof : 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
squash: ↓T
, 
exists: ∃x:A. B[x]
, 
so_lambda: λ2x.t[x]
, 
subtype_rel: A ⊆r B
, 
uimplies: b supposing a
, 
so_apply: x[s]
, 
implies: P 
⇒ Q
, 
bag-subtract: bag-subtract(eq;bs;as)
, 
all: ∀x:A. B[x]
, 
bag-accum: bag-accum(v,x.f[v; x];init;bs)
, 
list_accum: list_accum, 
nil: []
, 
it: ⋅
, 
bag-append: as + bs
, 
append: as @ bs
, 
list_ind: list_ind, 
top: Top
, 
so_lambda: λ2x y.t[x; y]
, 
so_apply: x[s1;s2]
, 
so_lambda: so_lambda(x,y,z.t[x; y; z])
, 
so_apply: x[s1;s2;s3]
, 
prop: ℙ
, 
true: True
, 
guard: {T}
, 
iff: P 
⇐⇒ Q
, 
and: P ∧ Q
, 
rev_implies: P 
⇐ Q
Lemmas referenced : 
bag_to_squash_list, 
list_induction, 
all_wf, 
list_wf, 
equal_wf, 
bag_wf, 
bag-subtract_wf, 
bag-append_wf, 
list-subtype-bag, 
list_accum_cons_lemma, 
list_ind_cons_lemma, 
bag-drop-head, 
append_wf, 
squash_wf, 
true_wf, 
iff_weakening_equal, 
deq_wf
Rules used in proof : 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
sqequalReflexivity, 
isect_memberFormation, 
introduction, 
cut, 
extract_by_obid, 
sqequalHypSubstitution, 
isectElimination, 
thin, 
hypothesisEquality, 
imageElimination, 
productElimination, 
promote_hyp, 
hypothesis, 
rename, 
sqequalRule, 
lambdaEquality, 
cumulativity, 
applyEquality, 
because_Cache, 
independent_isectElimination, 
independent_functionElimination, 
lambdaFormation, 
dependent_functionElimination, 
isect_memberEquality, 
voidElimination, 
voidEquality, 
equalityTransitivity, 
equalitySymmetry, 
natural_numberEquality, 
imageMemberEquality, 
baseClosed, 
universeEquality, 
hyp_replacement, 
applyLambdaEquality, 
axiomEquality
Latex:
\mforall{}[T:Type].  \mforall{}[eq:EqDecider(T)].  \mforall{}[as,bs:bag(T)].    (bag-subtract(eq;as  +  bs;as)  =  bs)
Date html generated:
2018_05_21-PM-09_49_25
Last ObjectModification:
2017_07_26-PM-06_31_03
Theory : bags_2
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