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))
Proof
Definitions occuring in Statement :
sqntype: sqntype(n;T),
nat: ℕ,
uimplies: b 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: n + m,
natural_number: $n,
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
uimplies: b supposing a,
sqntype: sqntype(n;T),
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
so_apply: x[s],
prop: ℙ,
pi1: fst(t),
pi2: snd(t),
nat: ℕ,
decidable: Dec(P),
or: P ∨ Q,
sq_stable: SqStable(P),
squash: ↓T,
uiff: uiff(P;Q),
and: P ∧ Q,
subtract: n - m,
subtype_rel: A ⊆r 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
Latex:
\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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