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:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(x1))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(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:

01(1(0(1(x1)))) → 01(0(1(0(x1))))
01(1(0(1(x1)))) → 01(x1)
01(0(0(0(x1)))) → 01(1(1(x1)))
01(1(0(1(x1)))) → 01(1(0(x1)))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(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:

01(1(0(1(x1)))) → 01(0(1(0(x1))))
01(1(0(1(x1)))) → 01(x1)
01(0(0(0(x1)))) → 01(1(1(x1)))
01(1(0(1(x1)))) → 01(1(0(x1)))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(x1))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

01(1(0(1(x1)))) → 01(0(1(0(x1))))
01(1(0(1(x1)))) → 01(x1)
01(1(0(1(x1)))) → 01(1(0(x1)))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(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.


01(1(0(1(x1)))) → 01(x1)
01(1(0(1(x1)))) → 01(1(0(x1)))
The remaining pairs can at least be oriented weakly.

01(1(0(1(x1)))) → 01(0(1(0(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 2 + (4)x_1   
POL(01(x1)) = (1/4)x_1   
POL(0(x1)) = 2 + (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(x1))))



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

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

01(1(0(1(x1)))) → 01(0(1(0(x1))))

The TRS R consists of the following rules:

0(0(0(0(x1)))) → 1(0(1(1(x1))))
0(1(0(1(x1)))) → 0(0(1(0(x1))))

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