### Nuprl Lemma : product_subtype_base

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

Proof

Definitions occuring in Statement :  uimplies: `b supposing a` subtype_rel: `A ⊆r 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: `b supposing a` subtype_rel: `A ⊆r B` all: `∀x:A. B[x]` so_apply: `x[s]` prop: `ℙ` 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

Latex:
\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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