Nuprl Lemma : sqn+1type_dep_product

[T:Type]. ∀[S:T ⟶ Type]. ∀[n:ℕ].  (sqntype(n 1;t:T × S[t])) supposing ((∀t:T. sqntype(n;S[t])) and sqntype(n;T))


Definitions occuring in Statement :  sqntype: sqntype(n;T) nat: uimplies: supposing a uall: [x:A]. B[x] so_apply: x[s] all: x:A. B[x] function: x:A ⟶ B[x] product: x:A × B[x] add: m natural_number: $n universe: Type
Definitions unfolded in proof :  uall: [x:A]. B[x] uimplies: supposing a sqntype: sqntype(n;T) all: x:A. B[x] implies:  Q member: t ∈ T so_apply: x[s] prop: pi1: fst(t) pi2: snd(t) nat: decidable: Dec(P) or: P ∨ Q sq_stable: SqStable(P) squash: T uiff: uiff(P;Q) and: P ∧ Q subtract: m subtype_rel: A ⊆B top: Top le: A ≤ B not: ¬A less_than': less_than'(a;b) true: True false: False guard: {T}
Lemmas referenced :  base_wf sqntype_wf nat_wf decidable__lt add-subtract-cancel sq_stable__le not-lt-2 condition-implies-le minus-add istype-void istype-int minus-one-mul zero-add minus-one-mul-top add-commutes add_functionality_wrt_le add-associates add-zero le-add-cancel
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity Error :isect_memberFormation_alt,  sqequalHypSubstitution sqequalRule Error :lambdaFormation_alt,  introduction Error :axiomSqequalN,  Error :equalityIsType4,  Error :productIsType,  Error :universeIsType,  hypothesisEquality applyEquality Error :inhabitedIsType,  cut extract_by_obid hypothesis Error :functionIsType,  isectElimination thin universeEquality equalityTransitivity equalitySymmetry productElimination Error :equalityIsType1,  dependent_functionElimination independent_functionElimination applyLambdaEquality setElimination rename natural_numberEquality because_Cache addEquality unionElimination sqequal_n rule imageMemberEquality baseClosed imageElimination independent_isectElimination Error :lambdaEquality_alt,  Error :isect_memberEquality_alt,  voidElimination minusEquality sqequalZero baseApply closedConclusion

\mforall{}[T:Type].  \mforall{}[S:T  {}\mrightarrow{}  Type].  \mforall{}[n:\mBbbN{}].
    (sqntype(n  +  1;t:T  \mtimes{}  S[t]))  supposing  ((\mforall{}t:T.  sqntype(n;S[t]))  and  sqntype(n;T))

Date html generated: 2019_06_20-AM-11_34_00
Last ObjectModification: 2018_09_29-PM-10_42_41

Theory : int_1

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