Nuprl Lemma : base-member-per-function

[A:Type]. ∀[B:per-function(A;x.Type)]. ∀[f:Base].
  f ∈ per-function(A;x.B[x]) supposing ∀[a,a':Base].  (f a) (f a') ∈ B[a] supposing a' ∈ A


Proof




Definitions occuring in Statement :  per-function: per-function(A;a.B[a]) uimplies: supposing a uall: [x:A]. B[x] so_apply: x[s] member: t ∈ T apply: a base: Base universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] member: t ∈ T type-function: type-function{i:l}(A) uimplies: supposing a function-eq: function-eq(A;a.B[a];f;g) per-function: per-function(A;a.B[a])
Lemmas referenced :  per-function_wf_type per-function-type-apply per-function_wf istype-universe base_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity Error :isect_memberFormation_alt,  cut introduction extract_by_obid sqequalHypSubstitution isectElimination thin hypothesisEquality because_Cache sqequalRule Error :isectIsType,  Error :inhabitedIsType,  Error :equalityIsType4,  hypothesis equalityTransitivity equalitySymmetry baseApply closedConclusion baseClosed Error :universeIsType,  universeEquality pertypeMemberEquality

Latex:
\mforall{}[A:Type].  \mforall{}[B:per-function(A;x.Type)].  \mforall{}[f:Base].
    f  \mmember{}  per-function(A;x.B[x])  supposing  \mforall{}[a,a':Base].    (f  a)  =  (f  a')  supposing  a  =  a'



Date html generated: 2019_06_20-AM-11_30_03
Last ObjectModification: 2018_10_06-AM-09_00_31

Theory : per!type


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