Nuprl Lemma : fun_exp_add_apply
∀[T:Type]. ∀[n,m:ℕ]. ∀[f:T ⟶ T]. ∀[x:T]. ((f^n (f^m x)) = (f^n + m 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,
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
member: t ∈ T,
squash: ↓T,
uall: ∀[x:A]. B[x],
prop: ℙ,
so_lambda: λ2x.t[x],
true: True,
so_apply: x[s],
subtype_rel: A ⊆r B,
uimplies: b supposing a,
guard: {T},
iff: P ⇐⇒ Q,
and: P ∧ Q,
rev_implies: P ⇐ Q,
implies: P ⇒ Q,
fun_exp: f^n,
primrec: primrec(n;b;c),
compose: f o g
Lemmas referenced :
compose_wf,
iff_weakening_equal,
fun_exp_add,
fun_exp_wf,
equal_wf,
nat_wf,
true_wf,
squash_wf,
uall_wf
Rules used in proof :
cut,
applyEquality,
instantiate,
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
lambdaEquality,
sqequalHypSubstitution,
imageElimination,
lemma_by_obid,
isectElimination,
thin,
hypothesisEquality,
equalityTransitivity,
hypothesis,
equalitySymmetry,
functionEquality,
cumulativity,
universeEquality,
because_Cache,
sqequalRule,
natural_numberEquality,
imageMemberEquality,
baseClosed,
independent_isectElimination,
productElimination,
independent_functionElimination,
isect_memberFormation,
introduction,
isect_memberEquality,
axiomEquality
Latex:
\mforall{}[T:Type]. \mforall{}[n,m:\mBbbN{}]. \mforall{}[f:T {}\mrightarrow{} T]. \mforall{}[x:T]. ((f\^{}n (f\^{}m x)) = (f\^{}n + m x))
Date html generated:
2016_05_13-PM-04_06_42
Last ObjectModification:
2016_01_14-PM-07_46_27
Theory : fun_1
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