Nuprl Lemma : sqntype_subtype
∀[A,B:Type]. ∀[n:ℕ]. (sqntype(n;A)) supposing (sqntype(n;B) and (A ⊆r B))
Proof
Definitions occuring in Statement :
sqntype: sqntype(n;T),
nat: ℕ,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
sqntype: sqntype(n;T),
all: ∀x:A. B[x],
implies: P ⇒ Q,
guard: {T},
prop: ℙ
Lemmas referenced :
equal_functionality_wrt_subtype_rel2,
istype-universe,
base_wf,
sqntype_wf,
subtype_rel_wf,
nat_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
introduction,
cut,
sqequalHypSubstitution,
Error :lambdaFormation_alt,
hypothesis,
dependent_functionElimination,
thin,
hypothesisEquality,
independent_functionElimination,
extract_by_obid,
isectElimination,
equalityTransitivity,
equalitySymmetry,
independent_isectElimination,
Error :equalityIsType4,
because_Cache,
Error :inhabitedIsType,
Error :universeIsType,
sqequalRule,
Error :lambdaEquality_alt,
Error :axiomSqequalN,
Error :functionIsTypeImplies,
Error :isect_memberEquality_alt,
universeEquality
Latex:
\mforall{}[A,B:Type]. \mforall{}[n:\mBbbN{}]. (sqntype(n;A)) supposing (sqntype(n;B) and (A \msubseteq{}r B))
Date html generated:
2019_06_20-AM-11_34_07
Last ObjectModification:
2018_10_06-AM-11_20_16
Theory : int_1
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