Nuprl Lemma : inl-one-one'
∀[A,B:Type]. ∀[x,y:A]. x = y ∈ A supposing (inl x) = (inl y) ∈ (A + B)
Proof
Definitions occuring in Statement :
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
inl: inl x,
union: left + right,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a
Lemmas referenced :
istype-universe
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation_alt,
introduction,
cut,
sqequalHypSubstitution,
sqequalRule,
applyLambdaEquality,
unionElimination,
thin,
hypothesisEquality,
hypothesis,
equalityIstype,
unionIsType,
universeIsType,
inlEquality_alt,
isect_memberEquality_alt,
isectElimination,
axiomEquality,
isectIsTypeImplies,
inhabitedIsType,
instantiate,
extract_by_obid,
universeEquality
Latex:
\mforall{}[A,B:Type]. \mforall{}[x,y:A]. x = y supposing (inl x) = (inl y)
Date html generated:
2020_05_19-PM-09_35_13
Last ObjectModification:
2019_12_05-PM-02_55_57
Theory : union
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