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:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(x1)))))))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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:

J(0(x1)) → P(p(s(s(0(p(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
I(0(x1)) → P(s(p(s(x1))))
I(s(x1)) → P(p(p(p(s(s(s(s(x1))))))))
J(0(x1)) → P(s(x1))
J(0(x1)) → P(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
I(s(x1)) → P(s(p(p(p(p(s(s(s(s(x1))))))))))
I(s(x1)) → P(p(s(s(s(s(x1))))))
J(s(x1)) → P(s(x1))
J(s(x1)) → I(p(s(p(s(x1)))))
I(0(x1)) → P(s(p(s(0(p(s(p(s(x1)))))))))
I(s(x1)) → P(s(s(s(s(x1)))))
I(s(x1)) → J(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))
J(s(x1)) → P(s(p(s(x1))))
J(s(x1)) → P(p(s(s(i(p(s(p(s(x1)))))))))
I(0(x1)) → P(s(0(p(s(p(s(x1)))))))
I(s(x1)) → P(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))
I(s(x1)) → P(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))
I(s(x1)) → P(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
J(0(x1)) → P(s(s(0(p(s(p(s(x1))))))))
I(0(x1)) → P(s(x1))
J(0(x1)) → P(s(p(s(x1))))
I(s(x1)) → P(p(p(s(s(s(s(x1)))))))
J(s(x1)) → P(s(s(i(p(s(p(s(x1))))))))

The TRS R consists of the following rules:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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:

J(0(x1)) → P(p(s(s(0(p(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
I(0(x1)) → P(s(p(s(x1))))
I(s(x1)) → P(p(p(p(s(s(s(s(x1))))))))
J(0(x1)) → P(s(x1))
J(0(x1)) → P(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
I(s(x1)) → P(s(p(p(p(p(s(s(s(s(x1))))))))))
I(s(x1)) → P(p(s(s(s(s(x1))))))
J(s(x1)) → P(s(x1))
J(s(x1)) → I(p(s(p(s(x1)))))
I(0(x1)) → P(s(p(s(0(p(s(p(s(x1)))))))))
I(s(x1)) → P(s(s(s(s(x1)))))
I(s(x1)) → J(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))
J(s(x1)) → P(s(p(s(x1))))
J(s(x1)) → P(p(s(s(i(p(s(p(s(x1)))))))))
I(0(x1)) → P(s(0(p(s(p(s(x1)))))))
I(s(x1)) → P(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))
I(s(x1)) → P(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))
I(s(x1)) → P(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
J(0(x1)) → P(s(s(0(p(s(p(s(x1))))))))
I(0(x1)) → P(s(x1))
J(0(x1)) → P(s(p(s(x1))))
I(s(x1)) → P(p(p(s(s(s(s(x1)))))))
J(s(x1)) → P(s(s(i(p(s(p(s(x1))))))))

The TRS R consists of the following rules:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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 21 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:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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)) = 2 + (2)x_1   
POL(s(x1)) = x_1   
The value of delta used in the strict ordering is 4.
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:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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:

J(s(x1)) → I(p(s(p(s(x1)))))
I(s(x1)) → J(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

i(0(x1)) → p(s(p(s(0(p(s(p(s(x1)))))))))
i(s(x1)) → p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))))))))
j(0(x1)) → p(s(p(p(s(s(0(p(s(p(s(x1)))))))))))
j(s(x1)) → s(s(s(s(p(p(s(s(i(p(s(p(s(x1)))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(x1)))))))))

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