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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