Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(x1))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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(2(0(x1))) → 11(x1)
01(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
L1(2(1(x1))) → 01(2(x1))
01(2(R(x1))) → 11(0(1(R(x1))))
11(2(1(x1))) → 01(2(x1))
L1(2(1(x1))) → L1(1(0(2(x1))))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))
01(2(R(x1))) → 01(1(R(x1)))
L1(2(0(x1))) → 11(0(1(x1)))
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
L1(2(1(x1))) → 11(0(2(x1)))
11(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → 01(1(x1))
11(2(R(x1))) → 01(1(R(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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(2(0(x1))) → 11(x1)
01(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
L1(2(1(x1))) → 01(2(x1))
01(2(R(x1))) → 11(0(1(R(x1))))
11(2(1(x1))) → 01(2(x1))
L1(2(1(x1))) → L1(1(0(2(x1))))
01(2(0(x1))) → 11(0(1(x1)))
01(2(0(x1))) → 01(1(x1))
01(2(R(x1))) → 01(1(R(x1)))
L1(2(0(x1))) → 11(0(1(x1)))
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
L1(2(1(x1))) → 11(0(2(x1)))
11(2(R(x1))) → 11(R(x1))
L1(2(0(x1))) → 01(1(x1))
11(2(R(x1))) → 01(1(R(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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 10 less nodes.

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

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

01(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
11(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 01(1(x1))
01(2(0(x1))) → 11(0(1(x1)))
01(2(1(x1))) → 01(2(x1))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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(2(0(x1))) → 11(x1)
11(2(0(x1))) → 01(1(x1))
11(2(0(x1))) → 11(x1)
01(2(1(x1))) → 11(0(2(x1)))
11(2(1(x1))) → 01(2(x1))
01(2(0(x1))) → 01(1(x1))
01(2(0(x1))) → 11(0(1(x1)))
01(2(1(x1))) → 01(2(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = 1 + (1/2)x_1   
POL(1(x1)) = 1 + (3/2)x_1   
POL(01(x1)) = 2 + (1/4)x_1   
POL(2(x1)) = 7/4 + (3)x_1   
POL(0(x1)) = 1/2 + (3/4)x_1   
POL(R(x1)) = 15/4   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] were oriented:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
1(2(0(x1))) → 2(0(1(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(x1))))



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

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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

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

L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))

The TRS R consists of the following rules:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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.


L1(2(0(x1))) → L1(1(0(1(x1))))
L1(2(1(x1))) → L1(1(0(2(x1))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = (4)x_1   
POL(L1(x1)) = (4)x_1   
POL(2(x1)) = 1/4   
POL(0(x1)) = 0   
POL(R(x1)) = 4 + (15/4)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
1(2(0(x1))) → 2(0(1(x1)))
0(2(0(x1))) → 1(0(1(x1)))
0(2(R(x1))) → 1(0(1(R(x1))))



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

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

1(2(1(x1))) → 2(0(2(x1)))
0(2(1(x1))) → 1(0(2(x1)))
L(2(1(x1))) → L(1(0(2(x1))))
1(2(0(x1))) → 2(0(1(x1)))
1(2(R(x1))) → 2(0(1(R(x1))))
0(2(0(x1))) → 1(0(1(x1)))
L(2(0(x1))) → L(1(0(1(x1))))
0(2(R(x1))) → 1(0(1(R(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.