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:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(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:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))
L1(f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(0), nil)
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
L1(f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
F(x, f(s(s(y)), nil)) → F(y, f(s(0), nil))
L1(f(s(s(y)), f(z, w))) → F(s(0), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, f(s(s(y)), f(z, w))) → F(y, f(s(z), w))
F(x, f(s(s(y)), nil)) → F(s(x), f(y, f(s(0), nil)))
F(x, f(s(s(y)), f(z, w))) → F(s(z), w)
F(x, f(s(s(y)), f(z, w))) → F(s(x), f(y, f(s(z), w)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = x_1 + x_2   
POL(s(x1)) = 1/4 + x_1   
POL(F(x1, x2)) = (1/4)x_2   
POL(0) = 0   
POL(nil) = 0   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))
f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(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:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(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:

L1(f(s(s(y)), f(z, w))) → L1(f(s(0), f(y, f(s(z), w))))

The TRS R consists of the following rules:

f(x, f(s(s(y)), f(z, w))) → f(s(x), f(y, f(s(z), w)))
L(f(s(s(y)), f(z, w))) → L(f(s(0), f(y, f(s(z), w))))
f(x, f(s(s(y)), nil)) → f(s(x), f(y, f(s(0), nil)))

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