Nuprl Lemma : strong-continuity2-half-squash-surject-biject-ext
∀[T,S,U:Type].
((U ⊆r ℕ)
⇒ (∃r:ℕ ⟶ U. ∀x:U. ((r x) = x ∈ U))
⇒ (∃g:ℕ ⟶ T. Surj(ℕ;T;g))
⇒ (∃h:S ⟶ U. Bij(S;U;h))
⇒ (∀F:(ℕ ⟶ T) ⟶ S
⇃(∃M:n:ℕ ⟶ (ℕn ⟶ T) ⟶ (S?)
∀f:ℕ ⟶ T
((∃n:ℕ. ((M n f) = (inl (F f)) ∈ (S?))) ∧ (∀n:ℕ. (M n f) = (inl (F f)) ∈ (S?) supposing ↑isl(M n f))))))
Proof
Definitions occuring in Statement :
biject: Bij(A;B;f),
surject: Surj(A;B;f),
quotient: x,y:A//B[x; y],
int_seg: {i..j-},
nat: ℕ,
assert: ↑b,
isl: isl(x),
uimplies: b supposing a,
subtype_rel: A ⊆r B,
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
exists: ∃x:A. B[x],
implies: P ⇒ Q,
and: P ∧ Q,
true: True,
unit: Unit,
apply: f a,
function: x:A ⟶ B[x],
inl: inl x,
union: left + right,
natural_number: $n,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
member: t ∈ T,
compose: f o g,
bfalse: ff,
it: ⋅,
pi1: fst(t),
strong-continuity-test: strong-continuity-test(M;n;f;b),
ifthenelse: if b then t else f fi ,
isl: isl(x),
btrue: tt,
eq_int: (i =z j),
subtract: n - m,
spreadn: spread3,
let: let,
strong-continuity2-half-squash-surject-biject,
implies-quotient-true2,
trivial-quotient-true,
strong-continuity2_biject_retract-ext,
strong-continuity2_functionality_surject,
strong-continuity2-half-squash,
implies-quotient-true,
strong-continuity2-iff-3,
strong-continuity3_functionality_surject,
basic-implies-strong-continuity2-ext,
strong-continuity2-implies-3,
surject-inverse,
decidable__assert,
strong-continuity-test-prop1,
strong-continuity-test-prop2,
not-isl-assert-isr,
any: any x,
bool_cases,
eqtt_to_assert,
uall: ∀[x:A]. B[x],
so_lambda: so_lambda(x,y,z,w.t[x; y; z; w]),
so_apply: x[s1;s2;s3;s4],
so_lambda: λ2x.t[x],
top: Top,
so_apply: x[s],
uimplies: b supposing a,
so_lambda: λ2x y.t[x; y],
so_apply: x[s1;s2],
all: ∀x:A. B[x],
implies: P ⇒ Q,
has-value: (a)↓
Lemmas referenced :
strong-continuity2-half-squash-surject-biject,
lifting-strict-decide,
istype-void,
strict4-decide,
lifting-strict-spread,
strict4-spread,
lifting-strict-callbyvalue,
lifting-strict-isint,
lifting-strict-int_eq,
has-value_wf_base,
is-exception_wf,
implies-quotient-true2,
trivial-quotient-true,
strong-continuity2_biject_retract-ext,
strong-continuity2_functionality_surject,
strong-continuity2-half-squash,
implies-quotient-true,
strong-continuity2-iff-3,
strong-continuity3_functionality_surject,
basic-implies-strong-continuity2-ext,
strong-continuity2-implies-3,
surject-inverse,
decidable__assert,
strong-continuity-test-prop1,
strong-continuity-test-prop2,
not-isl-assert-isr,
bool_cases,
eqtt_to_assert
Rules used in proof :
introduction,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
cut,
instantiate,
extract_by_obid,
hypothesis,
sqequalRule,
thin,
sqequalHypSubstitution,
equalityTransitivity,
equalitySymmetry,
isectElimination,
baseClosed,
Error :isect_memberEquality_alt,
voidElimination,
independent_isectElimination,
Error :inhabitedIsType,
Error :lambdaFormation_alt,
sqequalSqle,
divergentSqle,
callbyvalueDecide,
hypothesisEquality,
unionElimination,
sqleReflexivity,
Error :equalityIstype,
dependent_functionElimination,
independent_functionElimination,
decideExceptionCases,
axiomSqleEquality,
exceptionSqequal,
baseApply,
closedConclusion
Latex:
\mforall{}[T,S,U:Type].
((U \msubseteq{}r \mBbbN{})
{}\mRightarrow{} (\mexists{}r:\mBbbN{} {}\mrightarrow{} U. \mforall{}x:U. ((r x) = x))
{}\mRightarrow{} (\mexists{}g:\mBbbN{} {}\mrightarrow{} T. Surj(\mBbbN{};T;g))
{}\mRightarrow{} (\mexists{}h:S {}\mrightarrow{} U. Bij(S;U;h))
{}\mRightarrow{} (\mforall{}F:(\mBbbN{} {}\mrightarrow{} T) {}\mrightarrow{} S
\00D9(\mexists{}M:n:\mBbbN{} {}\mrightarrow{} (\mBbbN{}n {}\mrightarrow{} T) {}\mrightarrow{} (S?)
\mforall{}f:\mBbbN{} {}\mrightarrow{} T
((\mexists{}n:\mBbbN{}. ((M n f) = (inl (F f))))
\mwedge{} (\mforall{}n:\mBbbN{}. (M n f) = (inl (F f)) supposing \muparrow{}isl(M n f))))))
Date html generated:
2019_06_20-PM-02_51_25
Last ObjectModification:
2019_03_26-AM-06_46_54
Theory : continuity
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