Nuprl Lemma : sub-bag-filter3

T:Type. ∀p:T ⟶ 𝔹. ∀b,c:bag(T).  (sub-bag(T;b;c)  sub-bag(T;[x∈b|p[x]];[x∈c|p[x]]))


Proof




Definitions occuring in Statement :  sub-bag: sub-bag(T;as;bs) bag-filter: [x∈b|p[x]] bag: bag(T) bool: 𝔹 so_apply: x[s] all: x:A. B[x] implies:  Q function: x:A ⟶ B[x] universe: Type
Definitions unfolded in proof :  all: x:A. B[x] implies:  Q sub-bag: sub-bag(T;as;bs) exists: x:A. B[x] member: t ∈ T uall: [x:A]. B[x] so_lambda: λ2x.t[x] so_apply: x[s] subtype_rel: A ⊆B prop: uimplies: supposing a top: Top
Lemmas referenced :  bag-filter_wf subtype_rel_bag assert_wf equal_wf bag_wf bag-append_wf sub-bag_wf bool_wf bag-filter-append
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation sqequalHypSubstitution productElimination thin dependent_pairFormation cut introduction extract_by_obid isectElimination cumulativity hypothesisEquality sqequalRule lambdaEquality applyEquality functionExtensionality hypothesis setEquality independent_isectElimination setElimination rename because_Cache equalityTransitivity equalitySymmetry functionEquality universeEquality isect_memberEquality voidElimination voidEquality hyp_replacement Error :applyLambdaEquality

Latex:
\mforall{}T:Type.  \mforall{}p:T  {}\mrightarrow{}  \mBbbB{}.  \mforall{}b,c:bag(T).    (sub-bag(T;b;c)  {}\mRightarrow{}  sub-bag(T;[x\mmember{}b|p[x]];[x\mmember{}c|p[x]]))



Date html generated: 2016_10_25-AM-10_30_32
Last ObjectModification: 2016_07_12-AM-06_46_44

Theory : bags


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