Nuprl Lemma : bag-in-subtype2

[A,B:Type].  ∀[b:bag(B)]. b ∈ bag(A) supposing ∀x:B. (x ↓∈  (∃a:bag(A). x ↓∈ a)) supposing strong-subtype(A;B)


Proof




Definitions occuring in Statement :  bag-member: x ↓∈ bs bag: bag(T) strong-subtype: strong-subtype(A;B) uimplies: supposing a uall: [x:A]. B[x] all: x:A. B[x] exists: x:A. B[x] implies:  Q member: t ∈ T universe: Type
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a all: x:A. B[x] implies:  Q prop: exists: x:A. B[x] subtype_rel: A ⊆B strong-subtype: strong-subtype(A;B) cand: c∧ B
Lemmas referenced :  bag-in-subtype bag-member_wf bag_wf subtype_rel_bag strong-subtype_wf istype-universe bag-member-strong-subtype
Rules used in proof :  cut introduction extract_by_obid sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation_alt hypothesis sqequalHypSubstitution isectElimination thin hypothesisEquality independent_isectElimination lambdaFormation_alt dependent_functionElimination independent_functionElimination universeIsType equalityTransitivity equalitySymmetry sqequalRule axiomEquality functionIsType productIsType applyEquality productElimination isect_memberEquality_alt isectIsTypeImplies inhabitedIsType instantiate universeEquality

Latex:
\mforall{}[A,B:Type].
    \mforall{}[b:bag(B)].  b  \mmember{}  bag(A)  supposing  \mforall{}x:B.  (x  \mdownarrow{}\mmember{}  b  {}\mRightarrow{}  (\mexists{}a:bag(A).  x  \mdownarrow{}\mmember{}  a)) 
    supposing  strong-subtype(A;B)



Date html generated: 2019_10_15-AM-11_01_29
Last ObjectModification: 2019_08_15-PM-03_50_04

Theory : bags


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