Nuprl Lemma : bag-append-comm-assoc
∀[T:Type]. ∀[as,bs,cs:bag(T)]. ((as + bs + cs) = ((as + cs) + bs) ∈ bag(T))
Proof
Definitions occuring in Statement :
bag-append: as + bs,
bag: bag(T),
uall: ∀[x:A]. B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
top: Top,
prop: ℙ
Lemmas referenced :
bag-append-comm,
bag-append-assoc,
bag-append_wf,
equal_wf,
bag_wf
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isectElimination,
thin,
hypothesisEquality,
hypothesis,
sqequalRule,
isect_memberEquality,
voidElimination,
voidEquality,
cumulativity,
equalityTransitivity,
equalitySymmetry,
hyp_replacement,
Error :applyLambdaEquality,
because_Cache,
universeEquality,
isect_memberFormation,
axiomEquality
Latex:
\mforall{}[T:Type]. \mforall{}[as,bs,cs:bag(T)]. ((as + bs + cs) = ((as + cs) + bs))
Date html generated:
2016_10_25-AM-10_21_55
Last ObjectModification:
2016_07_12-AM-06_38_42
Theory : bags
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