### Nuprl Lemma : list-decomp-no_repeats

`∀[T:Type]. ∀[l1,l2,l3,l4:T List]. ∀[x:T].`
`  ((l1 = l3 ∈ (T List)) ∧ (l2 = l4 ∈ (T List))) supposing `
`     ((((l1 @ [x]) @ l2) = ((l3 @ [x]) @ l4) ∈ (T List)) and `
`     no_repeats(T;(l1 @ [x]) @ l2))`

Proof

Definitions occuring in Statement :  no_repeats: `no_repeats(T;l)` append: `as @ bs` cons: `[a / b]` nil: `[]` list: `T List` uimplies: `b supposing a` uall: `∀[x:A]. B[x]` and: `P ∧ Q` universe: `Type` equal: `s = t ∈ T`
Definitions unfolded in proof :  uall: `∀[x:A]. B[x]` member: `t ∈ T` uimplies: `b supposing a` and: `P ∧ Q` cand: `A c∧ B` prop: `ℙ` iff: `P `⇐⇒` Q` implies: `P `` Q` rev_implies: `P `` Q` all: `∀x:A. B[x]` subtype_rel: `A ⊆r B` le: `A ≤ B` less_than': `less_than'(a;b)` false: `False` not: `¬A` squash: `↓T` top: `Top` true: `True` guard: `{T}` int_seg: `{i..j-}` lelt: `i ≤ j < k` decidable: `Dec(P)` or: `P ∨ Q` uiff: `uiff(P;Q)` satisfiable_int_formula: `satisfiable_int_formula(fmla)` exists: `∃x:A. B[x]` ge: `i ≥ j ` sq_type: `SQType(T)` select: `L[n]` cons: `[a / b]` less_than: `a < b` no_repeats: `no_repeats(T;l)` nat: `ℕ` subtract: `n - m`
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation introduction cut independent_pairFormation hypothesis sqequalRule sqequalHypSubstitution productElimination thin independent_pairEquality axiomEquality extract_by_obid isectElimination cumulativity hypothesisEquality isect_memberEquality because_Cache equalityTransitivity equalitySymmetry universeEquality independent_functionElimination lambdaFormation dependent_functionElimination applyEquality natural_numberEquality addEquality independent_isectElimination lambdaEquality imageElimination intEquality voidElimination voidEquality imageMemberEquality baseClosed applyLambdaEquality setElimination rename unionElimination pointwiseFunctionality promote_hyp baseApply closedConclusion dependent_pairFormation int_eqEquality computeAll dependent_set_memberEquality instantiate hyp_replacement multiplyEquality productEquality

Latex:
\mforall{}[T:Type].  \mforall{}[l1,l2,l3,l4:T  List].  \mforall{}[x:T].
((l1  =  l3)  \mwedge{}  (l2  =  l4))  supposing
((((l1  @  [x])  @  l2)  =  ((l3  @  [x])  @  l4))  and
no\_repeats(T;(l1  @  [x])  @  l2))

Date html generated: 2017_10_01-AM-08_39_11
Last ObjectModification: 2017_07_26-PM-04_27_23

Theory : list!

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