Nuprl Lemma : bag-subtype-fset

[A:Type]. (bag(A) ⊆fset(A))


Proof




Definitions occuring in Statement :  bag: bag(T) fset: fset(T) subtype_rel: A ⊆B uall: [x:A]. B[x] universe: Type
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T subtype_rel: A ⊆B bag: bag(T) quotient: x,y:A//B[x; y] and: P ∧ Q fset: fset(T) so_lambda: λ2y.t[x; y] so_apply: x[s1;s2] uimplies: supposing a all: x:A. B[x] implies:  Q prop: guard: {T} set-equal: set-equal(T;x;y)
Lemmas referenced :  fset_wf quotient-member-eq list_wf set-equal_wf set-equal-equiv equal-wf-base permutation_wf bag_wf member-permutation
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut lambdaEquality sqequalHypSubstitution pointwiseFunctionalityForEquality extract_by_obid isectElimination thin cumulativity hypothesisEquality hypothesis sqequalRule pertypeElimination productElimination independent_isectElimination dependent_functionElimination equalityTransitivity equalitySymmetry independent_functionElimination productEquality because_Cache axiomEquality

Latex:
\mforall{}[A:Type].  (bag(A)  \msubseteq{}r  fset(A))



Date html generated: 2018_05_21-PM-06_24_04
Last ObjectModification: 2017_11_10-PM-04_45_12

Theory : bags


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