Nuprl Lemma : list-equal-subsume

[A,B:Type]. ∀[x,y:A List].  {x y ∈ (B List) supposing y ∈ (A List)} supposing {a:A| (a ∈ x)}  ⊆B


Proof




Definitions occuring in Statement :  l_member: (x ∈ l) list: List uimplies: supposing a subtype_rel: A ⊆B uall: [x:A]. B[x] guard: {T} set: {x:A| B[x]}  universe: Type equal: t ∈ T
Definitions unfolded in proof :  guard: {T} uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a prop: so_lambda: λ2x.t[x] so_apply: x[s] subtype_rel: A ⊆B
Lemmas referenced :  strong-subtype-equal-lists l_member_wf strong-subtype-set3 strong-subtype-self list-subtype subtype_rel_list equal_wf list_wf subtype_rel_wf
Rules used in proof :  sqequalSubstitution sqequalRule sqequalReflexivity sqequalTransitivity computationStep isect_memberFormation introduction cut lemma_by_obid sqequalHypSubstitution isectElimination thin setEquality hypothesisEquality hypothesis independent_isectElimination because_Cache lambdaEquality cumulativity equalityTransitivity equalitySymmetry applyEquality isect_memberEquality axiomEquality universeEquality

Latex:
\mforall{}[A,B:Type].  \mforall{}[x,y:A  List].    \{x  =  y  supposing  x  =  y\}  supposing  \{a:A|  (a  \mmember{}  x)\}    \msubseteq{}r  B



Date html generated: 2016_05_14-AM-07_40_51
Last ObjectModification: 2015_12_26-PM-02_51_27

Theory : list_1


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