The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
a__dbl1(0) → 01
a__dbl1(s(X)) → s1(s1(a__dbl1(mark(X))))
a__sel1(0, cons(X, Y)) → mark(X)
a__sel1(s(X), cons(Y, Z)) → a__sel1(mark(X), mark(Z))
a__quote(0) → 01
a__quote(s(X)) → s1(a__quote(mark(X)))
a__quote(dbl(X)) → a__dbl1(mark(X))
a__quote(sel(X, Y)) → a__sel1(mark(X), mark(Y))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(dbl1(X)) → a__dbl1(mark(X))
mark(sel1(X1, X2)) → a__sel1(mark(X1), mark(X2))
mark(quote(X)) → a__quote(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
mark(01) → 01
mark(s1(X)) → s1(mark(X))
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)
a__dbl1(X) → dbl1(X)
a__sel1(X1, X2) → sel1(X1, X2)
a__quote(X) → quote(X)

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

a__dbl(0) → 0
a__dbl(s(X)) → s(s(dbl(X)))
a__dbls(nil) → nil
a__dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
a__sel(0, cons(X, Y)) → mark(X)
a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__indx(nil, X) → nil
a__indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
a__from(X) → cons(X, from(s(X)))
a__dbl1(0) → 01
a__dbl1(s(X)) → s1(s1(a__dbl1(mark(X))))
a__sel1(0, cons(X, Y)) → mark(X)
a__sel1(s(X), cons(Y, Z)) → a__sel1(mark(X), mark(Z))
a__quote(0) → 01
a__quote(s(X)) → s1(a__quote(mark(X)))
a__quote(dbl(X)) → a__dbl1(mark(X))
a__quote(sel(X, Y)) → a__sel1(mark(X), mark(Y))
mark(dbl(X)) → a__dbl(mark(X))
mark(dbls(X)) → a__dbls(mark(X))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(indx(X1, X2)) → a__indx(mark(X1), X2)
mark(from(X)) → a__from(X)
mark(dbl1(X)) → a__dbl1(mark(X))
mark(sel1(X1, X2)) → a__sel1(mark(X1), mark(X2))
mark(quote(X)) → a__quote(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
mark(01) → 01
mark(s1(X)) → s1(mark(X))
a__dbl(X) → dbl(X)
a__dbls(X) → dbls(X)
a__sel(X1, X2) → sel(X1, X2)
a__indx(X1, X2) → indx(X1, X2)
a__from(X) → from(X)
a__dbl1(X) → dbl1(X)
a__sel1(X1, X2) → sel1(X1, X2)
a__quote(X) → quote(X)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
a__sel(s(s(X1652_3)), cons(Y, cons(X13613_3, X23614_3))) →⁺ a__sel(s(X1652_3), cons(X13613_3, X23614_3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [X1652_3 / s(X1652_3), X23614_3 / cons(X13613_3, X23614_3)].
The result substitution is [Y / X13613_3].

(4) BOUNDS(n^1, INF)