Nuprl Lemma : product_subtype_base

[A:Type]. ∀[B:A ⟶ Type].  ((a:A × B[a]) ⊆Base) supposing ((∀a:A. (B[a] ⊆Base)) and (A ⊆Base))


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

\mforall{}[A:Type].  \mforall{}[B:A  {}\mrightarrow{}  Type].
    ((a:A  \mtimes{}  B[a])  \msubseteq{}r  Base)  supposing  ((\mforall{}a:A.  (B[a]  \msubseteq{}r  Base))  and  (A  \msubseteq{}r  Base))

Date html generated: 2016_05_13-PM-03_19_23
Last ObjectModification: 2016_01_14-PM-04_32_02

Theory : subtype_0

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