Termination w.r.t. Q proof of /tmp/tmpy6SoQV/aprove02.srs

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:

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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:

V(s(x1)) → P(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))
V(0(x1)) → P(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
V(s(x1)) → P(s(p(s(x1))))
V(0(x1)) → P(s(s(s(s(s(x1))))))
W(s(x1)) → P(s(s(x1)))
W(s(x1)) → V(p(p(s(s(s(p(p(s(s(x1))))))))))
W(0(x1)) → P(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))
W(s(x1)) → P(s(s(v(p(p(s(s(s(p(p(s(s(x1)))))))))))))
V(0(x1)) → P(p(s(s(s(s(s(x1)))))))
W(s(x1)) → P(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))
V(0(x1)) → P(s(s(0(p(p(s(s(s(s(s(x1)))))))))))
W(0(x1)) → P(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))))))
V(s(x1)) → P(s(x1))
W(0(x1)) → P(s(s(0(s(s(s(s(s(s(x1))))))))))
W(s(x1)) → P(p(s(s(s(p(p(s(s(x1)))))))))
W(0(x1)) → P(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))
W(0(x1)) → P(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))
V(s(x1)) → P(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1)))))))))))))))))))
W(0(x1)) → P(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
V(s(x1)) → P(s(s(p(s(p(s(x1)))))))
P(p(s(x1))) → P(x1)
W(s(x1)) → P(s(s(s(p(p(s(s(x1))))))))
W(0(x1)) → P(p(s(s(0(s(s(s(s(s(s(x1)))))))))))
W(0(x1)) → P(p(p(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))))
W(s(x1)) → P(p(s(s(x1))))
V(s(x1)) → P(p(s(s(p(s(p(s(x1))))))))
W(0(x1)) → P(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))
V(s(x1)) → W(p(p(s(s(p(s(p(s(x1)))))))))
P(0(x1)) → P(s(x1))

The TRS R consists of the following rules:

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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:

V(s(x1)) → P(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))
V(0(x1)) → P(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
V(s(x1)) → P(s(p(s(x1))))
V(0(x1)) → P(s(s(s(s(s(x1))))))
W(s(x1)) → P(s(s(x1)))
W(s(x1)) → V(p(p(s(s(s(p(p(s(s(x1))))))))))
W(0(x1)) → P(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))
W(s(x1)) → P(s(s(v(p(p(s(s(s(p(p(s(s(x1)))))))))))))
V(0(x1)) → P(p(s(s(s(s(s(x1)))))))
W(s(x1)) → P(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))
V(0(x1)) → P(s(s(0(p(p(s(s(s(s(s(x1)))))))))))
W(0(x1)) → P(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))))))
V(s(x1)) → P(s(x1))
W(0(x1)) → P(s(s(0(s(s(s(s(s(s(x1))))))))))
W(s(x1)) → P(p(s(s(s(p(p(s(s(x1)))))))))
W(0(x1)) → P(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))
W(0(x1)) → P(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))
V(s(x1)) → P(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1)))))))))))))))))))
W(0(x1)) → P(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
V(s(x1)) → P(s(s(p(s(p(s(x1)))))))
P(p(s(x1))) → P(x1)
W(s(x1)) → P(s(s(s(p(p(s(s(x1))))))))
W(0(x1)) → P(p(s(s(0(s(s(s(s(s(s(x1)))))))))))
W(0(x1)) → P(p(p(p(p(s(s(0(s(s(s(s(s(s(x1))))))))))))))
W(s(x1)) → P(p(s(s(x1))))
V(s(x1)) → P(p(s(s(p(s(p(s(x1))))))))
W(0(x1)) → P(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))
V(s(x1)) → W(p(p(s(s(p(s(p(s(x1)))))))))
P(0(x1)) → P(s(x1))

The TRS R consists of the following rules:

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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 26 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

P(p(s(x1))) → P(x1)

The TRS R consists of the following rules:

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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.

P(p(s(x1))) → P(x1)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x₁)) = (2)x_1
POL(p(x₁)) = 1 + x_1
POL(s(x₁)) = x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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:

V(s(x1)) → W(p(p(s(s(p(s(p(s(x1)))))))))
W(s(x1)) → V(p(p(s(s(s(p(p(s(s(x1))))))))))

The TRS R consists of the following rules:

v(s(x1)) → s(p(p(s(s(s(s(s(s(s(s(w(p(p(s(s(p(s(p(s(x1))))))))))))))))))))
v(0(x1)) → p(p(s(s(0(p(p(s(s(s(s(s(x1))))))))))))
w(s(x1)) → s(s(s(s(s(s(p(p(s(s(v(p(p(s(s(s(p(p(s(s(x1))))))))))))))))))))
w(0(x1)) → p(s(p(p(p(p(p(p(p(p(s(s(0(s(s(s(s(s(s(x1)))))))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(p(s(x1))))))))))

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