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:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 1(0(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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:

51(9(x1)) → 51(x1)
31(8(x1)) → 71(x1)
91(x1) → 21(x1)
31(5(x1)) → 81(9(7(x1)))
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
81(8(4(x1))) → 91(x1)
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 51(0(2(x1)))
51(9(x1)) → 21(6(5(x1)))
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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:

51(9(x1)) → 51(x1)
31(8(x1)) → 71(x1)
91(x1) → 21(x1)
31(5(x1)) → 81(9(7(x1)))
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
81(8(4(x1))) → 91(x1)
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 51(0(2(x1)))
51(9(x1)) → 21(6(5(x1)))
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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 2 SCCs with 2 less nodes.

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

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

31(8(x1)) → 71(x1)
91(x1) → 21(x1)
31(5(x1)) → 81(9(7(x1)))
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
81(8(4(x1))) → 91(x1)
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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.


81(8(4(x1))) → 91(x1)
The remaining pairs can at least be oriented weakly.

31(8(x1)) → 71(x1)
91(x1) → 21(x1)
31(5(x1)) → 81(9(7(x1)))
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))
Used ordering: Polynomial interpretation [25,35]:

POL(5(x1)) = 2 + (2)x_1   
POL(31(x1)) = 4   
POL(9(x1)) = 2   
POL(4(x1)) = 0   
POL(8(x1)) = 4 + x_1   
POL(71(x1)) = 4   
POL(1(x1)) = 4 + (2)x_1   
POL(21(x1)) = 4   
POL(7(x1)) = 0   
POL(3(x1)) = (3)x_1   
POL(81(x1)) = (2)x_1   
POL(2(x1)) = 0   
POL(6(x1)) = 0   
POL(91(x1)) = 4   
POL(0(x1)) = (3)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

3(1(x1)) → 4(1(x1))
3(5(x1)) → 8(9(7(x1)))
8(4(x1)) → 6(x1)
3(9(x1)) → 9(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 3(2(3(x1)))
2(6(x1)) → 4(3(x1))
8(8(4(x1))) → 1(9(x1))
9(x1) → 5(0(2(x1)))
7(1(x1)) → 6(9(x1))



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

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

31(8(x1)) → 71(x1)
91(x1) → 21(x1)
31(5(x1)) → 81(9(7(x1)))
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ QDP
          ↳ QDP

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

31(8(x1)) → 71(x1)
91(x1) → 21(x1)
91(x1) → 21(3(x1))
31(8(x1)) → 31(2(7(x1)))
31(8(x1)) → 21(7(x1))
31(9(x1)) → 91(3(x1))
31(9(x1)) → 31(x1)
31(5(x1)) → 71(x1)
21(6(x1)) → 31(x1)
71(1(x1)) → 91(x1)
91(x1) → 31(2(3(x1)))
91(x1) → 31(x1)
31(5(x1)) → 91(7(x1))

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 1(0(x1))

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

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

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

51(9(x1)) → 51(x1)

The TRS R consists of the following rules:

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 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.


51(9(x1)) → 51(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(9(x1)) = 1 + (4)x_1   
POL(51(x1)) = (4)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
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

3(1(x1)) → 4(1(x1))
5(9(x1)) → 2(6(5(x1)))
3(5(x1)) → 8(9(7(x1)))
9(x1) → 3(2(3(x1)))
8(4(x1)) → 6(x1)
2(6(x1)) → 4(3(x1))
3(8(x1)) → 3(2(7(x1)))
9(x1) → 5(0(2(x1)))
8(8(4(x1))) → 1(9(x1))
7(1(x1)) → 6(9(x1))
3(9(x1)) → 9(3(x1))
7(5(x1)) → 1(0(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.