MAYBE

Proof:
ConCon could not decide if this system is quasi-decreasing.
\cite{O02}, p. 214, Proposition 7.2.50 does not apply.
System R transformed to U(R).
Call external tool:
ttt2 - trs 30
Input:
  fstsplit(0, x) -> nil
  fstsplit(s(n), nil) -> nil
  fstsplit(s(n), cons(h, t)) -> cons(h, fstsplit(n, t))
  sndsplit(0, x) -> x
  sndsplit(s(n), nil) -> nil
  sndsplit(s(n), cons(h, t)) -> sndsplit(n, t)
  empty(nil) -> true
  empty(cons(h, t)) -> false
  leq(0, m) -> true
  leq(s(n), 0) -> false
  leq(s(n), s(m)) -> leq(n, m)
  length(nil) -> 0
  length(cons(h, t)) -> s(length(t))
  app(nil, x) -> x
  app(cons(h, t), x) -> cons(h, app(t, x))
  map_f(pid, nil) -> nil
  map_f(pid, cons(h, t)) -> app(f(pid, h), map_f(pid, t))
  ?2(false, store, m) -> process(app(map_f(self, nil), sndsplit(m, store)), m)
  ?1(true, store, m) -> ?2(empty(fstsplit(m, store)), store, m)
  process(store, m) -> ?1(leq(m, length(store)), store, m)
  ?4(false, store, m) -> process(sndsplit(m, app(map_f(self, nil), store)), m)
  ?3(false, store, m) -> ?4(empty(fstsplit(m, app(map_f(self, nil), store))), store, m)
  process(store, m) -> ?3(leq(m, length(store)), store, m)

The external tool could not decide termination.