The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

a__sel(s(X), cons(Y, Z)) → a__sel(mark(X), mark(Z))
a__sel(0, cons(X, Z)) → mark(X)
a__first(0, Z) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__sel1(s(X), cons(Y, Z)) → a__sel1(mark(X), mark(Z))
a__sel1(0, cons(X, Z)) → a__quote(X)
a__first1(0, Z) → nil1
a__first1(s(X), cons(Y, Z)) → cons1(a__quote(Y), a__first1(mark(X), mark(Z)))
a__quote(0) → 01
a__quote1(cons(X, Z)) → cons1(a__quote(X), a__quote1(Z))
a__quote1(nil) → nil1
a__quote(s(X)) → s1(a__quote(X))
a__quote(sel(X, Z)) → a__sel1(mark(X), mark(Z))
a__quote1(first(X, Z)) → a__first1(mark(X), mark(Z))
a__unquote(01) → 0
a__unquote(s1(X)) → s(a__unquote(mark(X)))
a__unquote1(nil1) → nil
a__unquote1(cons1(X, Z)) → a__fcons(a__unquote(mark(X)), a__unquote1(mark(Z)))
a__fcons(X, Z) → cons(mark(X), Z)
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(sel1(X1, X2)) → a__sel1(mark(X1), mark(X2))
mark(quote(X)) → a__quote(X)
mark(first1(X1, X2)) → a__first1(mark(X1), mark(X2))
mark(quote1(X)) → a__quote1(X)
mark(unquote(X)) → a__unquote(mark(X))
mark(unquote1(X)) → a__unquote1(mark(X))
mark(fcons(X1, X2)) → a__fcons(mark(X1), mark(X2))
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(nil) → nil
mark(nil1) → nil1
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
mark(01) → 01
mark(s1(X)) → s1(mark(X))
a__sel(X1, X2) → sel(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)
a__sel1(X1, X2) → sel1(X1, X2)
a__quote(X) → quote(X)
a__first1(X1, X2) → first1(X1, X2)
a__quote1(X) → quote1(X)
a__unquote(X) → unquote(X)
a__unquote1(X) → unquote1(X)
a__fcons(X1, X2) → fcons(X1, X2)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
mark(sel(from(X125125_3), X2)) →⁺ a__sel(cons(mark(mark(X125125_3)), from(s(mark(X125125_3)))), mark(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0].
The pumping substitution is [X125125_3 / sel(from(X125125_3), X2)].
The result substitution is [ ].

The rewrite sequence
mark(sel(from(X125125_3), X2)) →⁺ a__sel(cons(mark(mark(X125125_3)), from(s(mark(X125125_3)))), mark(X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0,0].
The pumping substitution is [X125125_3 / sel(from(X125125_3), X2)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)