Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

average2(s1(x), y) -> average2(x, s1(y))
average2(x, s1(s1(s1(y)))) -> s1(average2(s1(x), y))
average2(0, 0) -> 0
average2(0, s1(0)) -> 0
average2(0, s1(s1(0))) -> s1(0)

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

average2(s1(x), y) -> average2(x, s1(y))
average2(x, s1(s1(s1(y)))) -> s1(average2(s1(x), y))
average2(0, 0) -> 0
average2(0, s1(0)) -> 0
average2(0, s1(s1(0))) -> s1(0)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AVERAGE2(x, s1(s1(s1(y)))) -> AVERAGE2(s1(x), y)
AVERAGE2(s1(x), y) -> AVERAGE2(x, s1(y))

The TRS R consists of the following rules:

average2(s1(x), y) -> average2(x, s1(y))
average2(x, s1(s1(s1(y)))) -> s1(average2(s1(x), y))
average2(0, 0) -> 0
average2(0, s1(0)) -> 0
average2(0, s1(s1(0))) -> s1(0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
QDP

Q DP problem:
The TRS P consists of the following rules:

AVERAGE2(x, s1(s1(s1(y)))) -> AVERAGE2(s1(x), y)
AVERAGE2(s1(x), y) -> AVERAGE2(x, s1(y))

The TRS R consists of the following rules:

average2(s1(x), y) -> average2(x, s1(y))
average2(x, s1(s1(s1(y)))) -> s1(average2(s1(x), y))
average2(0, 0) -> 0
average2(0, s1(0)) -> 0
average2(0, s1(s1(0))) -> s1(0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.