Nuprl Lemma : termination-equality-base

[T:Type]. ∀[x,y:Base].  y ∈ supposing (x)↓ ∧ (x y ∈ partial(T)) supposing value-type(T)


Proof




Definitions occuring in Statement :  partial: partial(T) value-type: value-type(T) has-value: (a)↓ uimplies: supposing a uall: [x:A]. B[x] and: P ∧ Q base: Base universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] uimplies: supposing a and: P ∧ Q partial: partial(T) quotient: x,y:A//B[x; y] cand: c∧ B label: ...$L... t guard: {T} member: t ∈ T prop: per-partial: per-partial(T;x;y)
Lemmas referenced :  base-partial_wf per-partial_wf has-value_wf_base partial_wf istype-base value-type_wf istype-universe
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity Error :isect_memberFormation_alt,  sqequalHypSubstitution productElimination thin sqequalRule cut hypothesis pertypeElimination promote_hyp Error :productIsType,  Error :equalityIstype,  Error :universeIsType,  introduction extract_by_obid isectElimination hypothesisEquality sqequalBase equalitySymmetry because_Cache instantiate universeEquality independent_isectElimination

Latex:
\mforall{}[T:Type].  \mforall{}[x,y:Base].    x  =  y  supposing  (x)\mdownarrow{}  \mwedge{}  (x  =  y)  supposing  value-type(T)



Date html generated: 2019_06_20-PM-00_33_53
Last ObjectModification: 2018_12_22-PM-01_07_43

Theory : partial_1


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