Nuprl Lemma : subtype_rel_tunion

[A:Type]. ∀[B:A ⟶ Type]. ∀[C:Type]. ∀[D:C ⟶ Type].
  (⋃a:A.B[a] ⊆r ⋃c:C.D[c]) supposing ((∀a:A. (B[a] ⊆D[a])) and (A ⊆C))


Definitions occuring in Statement :  uimplies: supposing a subtype_rel: A ⊆B tunion: x:A.B[x] uall: [x:A]. B[x] so_apply: x[s] all: x:A. B[x] function: x:A ⟶ B[x] universe: Type
Definitions unfolded in proof :  tunion: x:A.B[x] uall: [x:A]. B[x] member: t ∈ T uimplies: supposing a so_apply: x[s] so_lambda: λ2x.t[x] subtype_rel: A ⊆B prop:
Lemmas referenced :  subtype_rel_wf all_wf subtype_rel_product subtype_rel_image
Rules used in proof :  sqequalSubstitution sqequalRule sqequalReflexivity sqequalTransitivity computationStep isect_memberFormation introduction cut lemma_by_obid sqequalHypSubstitution isectElimination thin productEquality hypothesisEquality applyEquality baseClosed independent_isectElimination lambdaEquality hypothesis axiomEquality isect_memberEquality because_Cache equalityTransitivity equalitySymmetry functionEquality cumulativity universeEquality

\mforall{}[A:Type].  \mforall{}[B:A  {}\mrightarrow{}  Type].  \mforall{}[C:Type].  \mforall{}[D:C  {}\mrightarrow{}  Type].
    (\mcup{}a:A.B[a]  \msubseteq{}r  \mcup{}c:C.D[c])  supposing  ((\mforall{}a:A.  (B[a]  \msubseteq{}r  D[a]))  and  (A  \msubseteq{}r  C))

Date html generated: 2016_05_13-PM-03_18_41
Last ObjectModification: 2016_01_14-PM-04_32_01

Theory : subtype_0

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