Nuprl Lemma : add-has-value-partial-nat
∀[x,y:partial(ℕ)]. {(x ∈ ℤ) ∧ (y ∈ ℤ)} supposing (x + y)↓
Proof
Definitions occuring in Statement :
partial: partial(T),
nat: ℕ,
has-value: (a)↓,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
guard: {T},
and: P ∧ Q,
member: t ∈ T,
add: n + m,
int: ℤ
Definitions unfolded in proof :
all: ∀x:A. B[x],
member: t ∈ T,
implies: P ⇒ Q,
nat: ℕ,
uall: ∀[x:A]. B[x],
so_lambda: λ2x.t[x],
so_apply: x[s],
uimplies: b supposing a,
guard: {T},
and: P ∧ Q,
subtype_rel: A ⊆r B,
prop: ℙ,
has-value: (a)↓
Lemmas referenced :
subtype_partial_sqtype_base,
nat_wf,
set_subtype_base,
le_wf,
istype-int,
int_subtype_base,
has-value_wf_base,
partial_wf
Rules used in proof :
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
dependent_functionElimination,
thin,
hypothesis,
independent_functionElimination,
sqequalRule,
isectElimination,
intEquality,
Error :lambdaEquality_alt,
natural_numberEquality,
hypothesisEquality,
independent_isectElimination,
Error :isect_memberFormation_alt,
productElimination,
independent_pairEquality,
axiomEquality,
equalityTransitivity,
equalitySymmetry,
Error :universeIsType,
baseApply,
closedConclusion,
baseClosed,
applyEquality,
Error :isect_memberEquality_alt,
because_Cache,
Error :inhabitedIsType,
callbyvalueAdd,
independent_pairFormation
Latex:
\mforall{}[x,y:partial(\mBbbN{})]. \{(x \mmember{} \mBbbZ{}) \mwedge{} (y \mmember{} \mBbbZ{})\} supposing (x + y)\mdownarrow{}
Date html generated:
2019_06_20-PM-00_34_44
Last ObjectModification:
2018_10_07-AM-00_23_36
Theory : partial_1
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