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:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(x1))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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:

11(1(h(b(x1)))) → 11(1(s(b(x1))))
11(1(h(b(x1)))) → 11(s(b(x1)))
B(t(x1)) → H(x1)
11(t(x1)) → 11(1(1(x1)))
11(s(x1)) → 11(x1)
B(s(x1)) → H(x1)
H(1(b(x1))) → 11(1(b(x1)))
H(1(1(x1))) → 11(h(x1))
B(s(x1)) → B(h(x1))
11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(x1)
H(1(1(x1))) → H(x1)
B(t(x1)) → B(h(x1))

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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:

11(1(h(b(x1)))) → 11(1(s(b(x1))))
11(1(h(b(x1)))) → 11(s(b(x1)))
B(t(x1)) → H(x1)
11(t(x1)) → 11(1(1(x1)))
11(s(x1)) → 11(x1)
B(s(x1)) → H(x1)
H(1(b(x1))) → 11(1(b(x1)))
H(1(1(x1))) → 11(h(x1))
B(s(x1)) → B(h(x1))
11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(x1)
H(1(1(x1))) → H(x1)
B(t(x1)) → B(h(x1))

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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 4 less nodes.

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

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

11(1(h(b(x1)))) → 11(1(s(b(x1))))
11(1(h(b(x1)))) → 11(s(b(x1)))
11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(1(1(x1)))
11(s(x1)) → 11(x1)
11(t(x1)) → 11(x1)

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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.


11(1(h(b(x1)))) → 11(1(s(b(x1))))
11(1(h(b(x1)))) → 11(s(b(x1)))
The remaining pairs can at least be oriented weakly.

11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(1(1(x1)))
11(s(x1)) → 11(x1)
11(t(x1)) → 11(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = (1/4)x_1   
POL(1(x1)) = (1/4)x_1   
POL(t(x1)) = x_1   
POL(h(x1)) = 3/4   
POL(s(x1)) = (2)x_1   
POL(b(x1)) = 0   
The value of delta used in the strict ordering is 3/64.
The following usable rules [17] were oriented:

1(s(x1)) → s(1(x1))
1(t(x1)) → t(1(1(1(x1))))
h(1(b(x1))) → t(1(1(b(x1))))
b(t(x1)) → b(h(x1))
b(s(x1)) → b(h(x1))
h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))



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

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

11(t(x1)) → 11(1(1(x1)))
11(t(x1)) → 11(1(x1))
11(s(x1)) → 11(x1)
11(t(x1)) → 11(x1)

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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.


11(s(x1)) → 11(x1)
The remaining pairs can at least be oriented weakly.

11(t(x1)) → 11(1(1(x1)))
11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = (1/4)x_1   
POL(1(x1)) = x_1   
POL(t(x1)) = (2)x_1   
POL(h(x1)) = 1/4 + (4)x_1   
POL(s(x1)) = 1/2 + (2)x_1   
POL(b(x1)) = 4   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

1(s(x1)) → s(1(x1))
1(t(x1)) → t(1(1(1(x1))))
h(1(b(x1))) → t(1(1(b(x1))))
b(t(x1)) → b(h(x1))
b(s(x1)) → b(h(x1))
h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))



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

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

11(t(x1)) → 11(1(x1))
11(t(x1)) → 11(1(1(x1)))
11(t(x1)) → 11(x1)

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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.


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

POL(11(x1)) = (1/2)x_1   
POL(1(x1)) = x_1   
POL(t(x1)) = 1/4 + (4)x_1   
POL(h(x1)) = 1/4   
POL(s(x1)) = 1/4   
POL(b(x1)) = 0   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

1(s(x1)) → s(1(x1))
1(t(x1)) → t(1(1(1(x1))))
h(1(b(x1))) → t(1(1(b(x1))))
b(t(x1)) → b(h(x1))
b(s(x1)) → b(h(x1))
h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))



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

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

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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:

H(1(1(x1))) → H(x1)

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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.


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

POL(1(x1)) = 1 + x_1   
POL(H(x1)) = (2)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
          ↳ QDP

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

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(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:

B(t(x1)) → B(h(x1))
B(s(x1)) → B(h(x1))

The TRS R consists of the following rules:

h(1(1(x1))) → 1(h(x1))
1(1(h(b(x1)))) → 1(1(s(b(x1))))
1(s(x1)) → s(1(x1))
b(s(x1)) → b(h(x1))
h(1(b(x1))) → t(1(1(b(x1))))
1(t(x1)) → t(1(1(1(x1))))
b(t(x1)) → b(h(x1))

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