Nuprl Lemma : list-equal-subsume
∀[A,B:Type]. ∀[x,y:A List]. {x = y ∈ (B List) supposing x = y ∈ (A List)} supposing {a:A| (a ∈ x)} ⊆r B
Proof
Definitions occuring in Statement :
l_member: (x ∈ l),
list: T List,
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
guard: {T},
set: {x:A| B[x]} ,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
guard: {T},
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
prop: ℙ,
so_lambda: λ2x.t[x],
so_apply: x[s],
subtype_rel: A ⊆r B
Lemmas referenced :
strong-subtype-equal-lists,
l_member_wf,
strong-subtype-set3,
strong-subtype-self,
list-subtype,
subtype_rel_list,
equal_wf,
list_wf,
subtype_rel_wf
Rules used in proof :
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
isect_memberFormation,
introduction,
cut,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
setEquality,
hypothesisEquality,
hypothesis,
independent_isectElimination,
because_Cache,
lambdaEquality,
cumulativity,
equalityTransitivity,
equalitySymmetry,
applyEquality,
isect_memberEquality,
axiomEquality,
universeEquality
Latex:
\mforall{}[A,B:Type]. \mforall{}[x,y:A List]. \{x = y supposing x = y\} supposing \{a:A| (a \mmember{} x)\} \msubseteq{}r B
Date html generated:
2016_05_14-AM-07_40_51
Last ObjectModification:
2015_12_26-PM-02_51_27
Theory : list_1
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