Nuprl Lemma : l_disjoint_member

[T:Type]. ∀[l1,l2:T List]. ∀[x:T].  (x ∈ l2)) supposing ((x ∈ l1) and l_disjoint(T;l1;l2))


Proof




Definitions occuring in Statement :  l_disjoint: l_disjoint(T;l1;l2) l_member: (x ∈ l) list: List uimplies: supposing a uall: [x:A]. B[x] not: ¬A universe: Type
Definitions unfolded in proof :  l_disjoint: l_disjoint(T;l1;l2) uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a not: ¬A implies:  Q false: False prop: all: x:A. B[x] and: P ∧ Q so_lambda: λ2x.t[x] so_apply: x[s] cand: c∧ B
Lemmas referenced :  l_member_wf not_wf all_wf list_wf
Rules used in proof :  sqequalSubstitution sqequalRule sqequalReflexivity sqequalTransitivity computationStep Error :isect_memberFormation_alt,  introduction cut lambdaFormation thin hypothesis sqequalHypSubstitution independent_functionElimination voidElimination extract_by_obid isectElimination hypothesisEquality lambdaEquality dependent_functionElimination because_Cache Error :universeIsType,  isect_memberEquality equalityTransitivity equalitySymmetry Error :functionIsType,  Error :inhabitedIsType,  productEquality universeEquality independent_pairFormation

Latex:
\mforall{}[T:Type].  \mforall{}[l1,l2:T  List].  \mforall{}[x:T].    (\mneg{}(x  \mmember{}  l2))  supposing  ((x  \mmember{}  l1)  and  l\_disjoint(T;l1;l2))



Date html generated: 2019_06_20-PM-01_26_57
Last ObjectModification: 2018_09_26-PM-05_37_07

Theory : list_1


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