Nuprl Lemma : fun_exp_add_apply1
∀[T:Type]. ∀[n:ℕ]. ∀[f:T ⟶ T]. ∀[x:T]. ((f^n (f x)) = (f^n + 1 x) ∈ T)
Proof
Definitions occuring in Statement :
fun_exp: f^n,
nat: ℕ,
uall: ∀[x:A]. B[x],
apply: f a,
function: x:A ⟶ B[x],
add: n + m,
natural_number: $n,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
nat: ℕ,
le: A ≤ B,
and: P ∧ Q,
less_than': less_than'(a;b),
false: False,
not: ¬A,
implies: P ⇒ Q,
prop: ℙ,
all: ∀x:A. B[x],
decidable: Dec(P),
or: P ∨ Q,
iff: P ⇐⇒ Q,
rev_implies: P ⇐ Q,
uiff: uiff(P;Q),
uimplies: b supposing a,
sq_stable: SqStable(P),
squash: ↓T,
subtract: n - m,
subtype_rel: A ⊆r B,
top: Top,
true: True,
guard: {T},
eq_int: (i =z j),
ifthenelse: if b then t else f fi ,
bfalse: ff,
compose: f o g,
btrue: tt
Lemmas referenced :
nat_wf,
false_wf,
le_wf,
fun_exp_wf,
decidable__le,
not-le-2,
sq_stable__le,
condition-implies-le,
minus-add,
minus-one-mul,
zero-add,
minus-one-mul-top,
add-associates,
add-swap,
add-commutes,
add_functionality_wrt_le,
add-zero,
le-add-cancel,
equal_wf,
squash_wf,
true_wf,
fun_exp_add_apply,
iff_weakening_equal,
fun_exp_unroll
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
hypothesis,
hypothesisEquality,
sqequalRule,
sqequalHypSubstitution,
isect_memberEquality,
isectElimination,
thin,
axiomEquality,
because_Cache,
functionEquality,
cumulativity,
extract_by_obid,
universeEquality,
dependent_set_memberEquality,
natural_numberEquality,
independent_pairFormation,
lambdaFormation,
applyEquality,
addEquality,
setElimination,
rename,
dependent_functionElimination,
unionElimination,
voidElimination,
productElimination,
independent_functionElimination,
independent_isectElimination,
imageMemberEquality,
baseClosed,
imageElimination,
lambdaEquality,
voidEquality,
intEquality,
minusEquality,
functionExtensionality,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[T:Type]. \mforall{}[n:\mBbbN{}]. \mforall{}[f:T {}\mrightarrow{} T]. \mforall{}[x:T]. ((f\^{}n (f x)) = (f\^{}n + 1 x))
Date html generated:
2017_04_14-AM-07_34_41
Last ObjectModification:
2017_02_27-PM-03_07_34
Theory : fun_1
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