Nuprl Lemma : union_subtype_base
∀[A,B:Type]. ((A + B) ⊆r Base) supposing ((B ⊆r Base) and (A ⊆r Base))
Proof
Definitions occuring in Statement :
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
union: left + right,
base: Base,
universe: Type
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
uimplies: b supposing a,
subtype_rel: A ⊆r B,
member: t ∈ T
Lemmas referenced :
istype-universe,
subtype_rel_wf,
base_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
Error :lambdaEquality_alt,
unionElimination,
thin,
sqequalRule,
baseApply,
closedConclusion,
baseClosed,
cut,
hypothesisEquality,
applyEquality,
hypothesis,
sqequalHypSubstitution,
Error :unionIsType,
introduction,
extract_by_obid,
isectElimination,
Error :universeIsType,
Error :inhabitedIsType,
universeEquality
Latex:
\mforall{}[A,B:Type]. ((A + B) \msubseteq{}r Base) supposing ((B \msubseteq{}r Base) and (A \msubseteq{}r Base))
Date html generated:
2019_06_20-AM-11_19_28
Last ObjectModification:
2018_10_06-AM-09_07_21
Theory : subtype_0
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