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:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → x1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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:

SQ(0(x1)) → P(p(p(s(s(s(s(0(p(s(p(s(x1))))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
SQ(0(x1)) → 01(p(s(p(s(x1)))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(0(x1)) → P(s(s(s(s(0(p(s(p(s(x1))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → 01(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))
SQ(0(x1)) → P(p(s(s(s(s(0(p(s(p(s(x1)))))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
P(0(x1)) → 01(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
SQ(0(x1)) → P(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
SQ(0(x1)) → P(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
SQ(0(x1)) → P(s(x1))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(0(x1)) → P(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))

The TRS R consists of the following rules:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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:

SQ(0(x1)) → P(p(p(s(s(s(s(0(p(s(p(s(x1))))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
SQ(0(x1)) → 01(p(s(p(s(x1)))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(0(x1)) → P(s(s(s(s(0(p(s(p(s(x1))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → 01(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))
SQ(0(x1)) → P(p(s(s(s(s(0(p(s(p(s(x1)))))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
P(0(x1)) → 01(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
SQ(0(x1)) → P(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
SQ(0(x1)) → P(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
SQ(0(x1)) → P(s(x1))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(0(x1)) → P(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))

The TRS R consists of the following rules:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → x1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 46 less nodes.

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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
          ↳ QDP

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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
            ↳ QDPOrderProof
          ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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.


TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 4 + (4)x_1   
POL(p(x1)) = 3/4 + (1/4)x_1   
POL(TWICE(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

p(s(x1)) → x1



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

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → 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
          ↳ QDP

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(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(s(s(s(x1))))))))))))
0(x1) → x1

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