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:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

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:

L11(f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
L11(f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(0), nil)
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(x), f2(y, f2(s1(0), nil)))
L11(f2(s1(s1(y)), f2(z, w))) -> L11(f2(s1(0), f2(y, f2(s1(z), w))))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(x), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))
L11(f2(s1(s1(y)), f2(z, w))) -> F2(s1(0), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), nil)) -> F2(y, f2(s1(0), nil))

The TRS R consists of the following rules:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

L11(f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
L11(f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(0), nil)
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(x), f2(y, f2(s1(0), nil)))
L11(f2(s1(s1(y)), f2(z, w))) -> L11(f2(s1(0), f2(y, f2(s1(z), w))))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(x), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))
L11(f2(s1(s1(y)), f2(z, w))) -> F2(s1(0), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), nil)) -> F2(y, f2(s1(0), nil))

The TRS R consists of the following rules:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(x), f2(y, f2(s1(0), nil)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(x), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))

The TRS R consists of the following rules:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(z), w)
F2(x, f2(s1(s1(y)), nil)) -> F2(s1(x), f2(y, f2(s1(0), nil)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(s1(x), f2(y, f2(s1(z), w)))
F2(x, f2(s1(s1(y)), f2(z, w))) -> F2(y, f2(s1(z), w))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(F2(x1, x2)) = x2   
POL(f2(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))
f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

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

L11(f2(s1(s1(y)), f2(z, w))) -> L11(f2(s1(0), f2(y, f2(s1(z), w))))

The TRS R consists of the following rules:

f2(x, f2(s1(s1(y)), f2(z, w))) -> f2(s1(x), f2(y, f2(s1(z), w)))
L1(f2(s1(s1(y)), f2(z, w))) -> L1(f2(s1(0), f2(y, f2(s1(z), w))))
f2(x, f2(s1(s1(y)), nil)) -> f2(s1(x), f2(y, f2(s1(0), nil)))

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