Nuprl Lemma : type-with-y=n
∀n,m:ℤ. ((¬(n = m ∈ ℤ))
⇒ (∀y:Base. ∃T:Type. ((y = n ∈ T) ∧ (m = m ∈ T) ∧ ((n = m ∈ T)
⇒ (y = m ∈ Base)))))
Proof
Definitions occuring in Statement :
all: ∀x:A. B[x]
,
exists: ∃x:A. B[x]
,
not: ¬A
,
implies: P
⇒ Q
,
and: P ∧ Q
,
int: ℤ
,
base: Base
,
universe: Type
,
equal: s = t ∈ T
Definitions unfolded in proof :
all: ∀x:A. B[x]
,
implies: P
⇒ Q
,
member: t ∈ T
,
not: ¬A
,
false: False
,
uimplies: b supposing a
,
so_apply: x[s]
,
so_lambda: λ2x.t[x]
,
uall: ∀[x:A]. B[x]
,
subtype_rel: A ⊆r B
,
or: P ∨ Q
,
cand: A c∧ B
,
and: P ∧ Q
,
refl: Refl(T;x,y.E[x; y])
,
sym: Sym(T;x,y.E[x; y])
,
trans: Trans(T;x,y.E[x; y])
,
equiv_rel: EquivRel(T;x,y.E[x; y])
,
exists: ∃x:A. B[x]
,
prop: ℙ
,
so_lambda: λ2x y.t[x; y]
,
so_apply: x[s1;s2]
,
quotient: x,y:A//B[x; y]
,
sq_type: SQType(T)
,
guard: {T}
Lemmas referenced :
istype-base,
istype-void,
istype-int,
subtype_rel_self,
int_subtype_base,
equal_wf,
or_wf,
base_wf,
set_subtype_base,
quotient_wf,
equal-wf-base,
quotient-member-eq,
subtype_base_sq
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :lambdaFormation_alt,
sqequalRule,
cut,
introduction,
extract_by_obid,
hypothesis,
Error :functionIsType,
Error :equalityIstype,
Error :inhabitedIsType,
hypothesisEquality,
independent_pairFormation,
Error :setIsType,
because_Cache,
Error :equalityIsType4,
Error :productIsType,
Error :unionIsType,
independent_isectElimination,
Error :lambdaEquality_alt,
thin,
isectElimination,
sqequalHypSubstitution,
applyEquality,
baseClosed,
closedConclusion,
baseApply,
Error :inlFormation_alt,
Error :inrFormation_alt,
productElimination,
equalityTransitivity,
equalitySymmetry,
unionElimination,
Error :dependent_pairFormation_alt,
setEquality,
unionEquality,
setElimination,
rename,
productEquality,
dependent_functionElimination,
Error :dependent_set_memberEquality_alt,
independent_functionElimination,
Error :universeIsType,
sqequalBase,
pertypeElimination,
promote_hyp,
instantiate,
cumulativity,
voidElimination
Latex:
\mforall{}n,m:\mBbbZ{}. ((\mneg{}(n = m)) {}\mRightarrow{} (\mforall{}y:Base. \mexists{}T:Type. ((y = n) \mwedge{} (m = m) \mwedge{} ((n = m) {}\mRightarrow{} (y = m)))))
Date html generated:
2019_06_20-PM-02_44_39
Last ObjectModification:
2018_12_06-PM-11_56_39
Theory : num_thy_1
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