Nuprl Lemma : fun-connected-induction
∀[T:Type]
∀f:T ⟶ T
∀[R:T ⟶ T ⟶ ℙ]
((∀x:T. R[x;x])
⇒ (∀x,y,z:T. (y is f*(z) ⇒ R[y;z] ⇒ R[x;z]) supposing ((¬(x = y ∈ T)) and (x = (f y) ∈ T)))
⇒ {∀x,y:T. (x is f*(y) ⇒ R[x;y])})
Proof
Definitions occuring in Statement :
fun-connected: y is f*(x),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
prop: ℙ,
guard: {T},
so_apply: x[s1;s2],
all: ∀x:A. B[x],
not: ¬A,
implies: P ⇒ Q,
apply: f a,
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
guard: {T},
uall: ∀[x:A]. B[x],
all: ∀x:A. B[x],
implies: P ⇒ Q,
member: t ∈ T,
prop: ℙ,
so_lambda: λ2x.t[x],
uimplies: b supposing a,
so_apply: x[s1;s2],
subtype_rel: A ⊆r B,
so_apply: x[s],
and: P ∧ Q,
not: ¬A,
fun-path: y=f*(x) via L,
fun-connected: y is f*(x),
exists: ∃x:A. B[x],
select: L[n],
nil: [],
it: ⋅,
so_lambda: λ2x y.t[x; y],
top: Top,
subtract: n - m,
less_than: a < b,
squash: ↓T,
less_than': less_than'(a;b),
false: False,
uiff: uiff(P;Q),
decidable: Dec(P),
or: P ∨ Q
Lemmas referenced :
all_wf,
isect_wf,
equal_wf,
not_wf,
fun-connected_wf,
fun-path_wf,
nil_wf,
less_than_wf,
length_wf,
cons_wf,
list_induction,
list_wf,
length_of_nil_lemma,
stuck-spread,
base_wf,
fun-path-cons,
decidable__lt
Rules used in proof :
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
isect_memberFormation,
lambdaFormation,
cut,
introduction,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
cumulativity,
hypothesisEquality,
lambdaEquality,
because_Cache,
applyEquality,
functionExtensionality,
hypothesis,
functionEquality,
universeEquality,
natural_numberEquality,
productEquality,
independent_isectElimination,
productElimination,
independent_functionElimination,
baseClosed,
isect_memberEquality,
voidElimination,
voidEquality,
imageElimination,
rename,
equalitySymmetry,
hyp_replacement,
applyLambdaEquality,
dependent_functionElimination,
unionElimination,
dependent_pairFormation
Latex:
\mforall{}[T:Type]
\mforall{}f:T {}\mrightarrow{} T
\mforall{}[R:T {}\mrightarrow{} T {}\mrightarrow{} \mBbbP{}]
((\mforall{}x:T. R[x;x])
{}\mRightarrow{} (\mforall{}x,y,z:T. (y is f*(z) {}\mRightarrow{} R[y;z] {}\mRightarrow{} R[x;z]) supposing ((\mneg{}(x = y)) and (x = (f y))))
{}\mRightarrow{} \{\mforall{}x,y:T. (x is f*(y) {}\mRightarrow{} R[x;y])\})
Date html generated:
2018_05_21-PM-07_44_06
Last ObjectModification:
2017_07_26-PM-05_21_51
Theory : general
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