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(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(x1)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q0(0(x1))) → 01(q1(x1))
H(q3(x1)) → 01(q3(x1))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q0(h(x1))) → 01(q0(h(x1)))
01(q0(h(x1))) → 01(0(q0(h(x1))))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
H(q1(x1)) → H(0(q1(x1)))
11(q3(x1)) → 11(x1)
01(q0(1(x1))) → 11(q0(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
H(q5(x1)) → H(0(q5(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(h(x1))) → 11(0(q5(h(x1))))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q0(1(x1))) → 11(q1(x1))
11(q4(x1)) → 11(x1)
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))
11(q1(0(x1))) → 01(q1(x1))
H(q3(x1)) → H(0(q3(x1)))
H(q0(x1)) → 01(q0(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q0(1(x1))) → 01(1(q0(x1)))
H(q4(x1)) → H(0(q4(x1)))
H(q4(x1)) → 01(q4(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 11(0(q1(h(x1))))
H(q2(x1)) → 01(q2(x1))
H(q1(x1)) → 01(q1(x1))
H(q0(x1)) → H(0(q0(x1)))
H(q5(x1)) → 01(q5(x1))
01(q4(0(x1))) → 11(0(q5(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
01(q1(1(x1))) → 11(q2(x1))
11(q5(0(x1))) → 01(q1(x1))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
01(q4(h(x1))) → 01(q5(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
H(q2(x1)) → H(0(q2(x1)))
01(q0(0(x1))) → 01(0(q0(x1)))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q0(0(x1))) → 01(q1(x1))
H(q3(x1)) → 01(q3(x1))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q0(h(x1))) → 01(q0(h(x1)))
01(q0(h(x1))) → 01(0(q0(h(x1))))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
H(q1(x1)) → H(0(q1(x1)))
11(q3(x1)) → 11(x1)
01(q0(1(x1))) → 11(q0(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
H(q5(x1)) → H(0(q5(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(h(x1))) → 11(0(q5(h(x1))))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q0(1(x1))) → 11(q1(x1))
11(q4(x1)) → 11(x1)
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))
11(q1(0(x1))) → 01(q1(x1))
H(q3(x1)) → H(0(q3(x1)))
H(q0(x1)) → 01(q0(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q0(1(x1))) → 01(1(q0(x1)))
H(q4(x1)) → H(0(q4(x1)))
H(q4(x1)) → 01(q4(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 11(0(q1(h(x1))))
H(q2(x1)) → 01(q2(x1))
H(q1(x1)) → 01(q1(x1))
H(q0(x1)) → H(0(q0(x1)))
H(q5(x1)) → 01(q5(x1))
01(q4(0(x1))) → 11(0(q5(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
01(q1(1(x1))) → 11(q2(x1))
11(q5(0(x1))) → 01(q1(x1))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
01(q4(h(x1))) → 01(q5(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
H(q2(x1)) → H(0(q2(x1)))
01(q0(0(x1))) → 01(0(q0(x1)))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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 14 less nodes.

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

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

11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q0(1(x1))) → 01(1(q0(x1)))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q0(h(x1))) → 01(q0(h(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
01(q0(h(x1))) → 01(0(q0(h(x1))))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q0(1(x1))) → 11(q0(x1))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q4(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))
01(q0(0(x1))) → 01(0(q0(x1)))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q0(1(x1))) → 01(1(q0(x1)))
01(q0(h(x1))) → 01(0(q0(h(x1))))
01(q0(1(x1))) → 11(q0(x1))
01(q0(0(x1))) → 01(0(q0(x1)))
The remaining pairs can at least be oriented weakly.

11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q0(h(x1))) → 01(q0(h(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q4(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(q1(x1)) = 0   
POL(11(x1)) = 0   
POL(1(x1)) = 0   
POL(01(x1)) = x_1   
POL(q0(x1)) = 2   
POL(q5(x1)) = 0   
POL(q4(x1)) = x_1   
POL(h(x1)) = 2   
POL(q2(x1)) = 0   
POL(0(x1)) = 0   
POL(q3(x1)) = x_1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
h(q4(x1)) → h(0(q4(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q5(x1)) → h(0(q5(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(0(x1))) → 0(0(q0(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(0(x1))) → 1(0(q2(x1)))
0(q4(1(x1))) → 1(1(q5(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
0(q1(h(x1))) → 0(0(q2(h(x1))))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))



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

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

11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q3(x1)) → 01(x1)
01(q1(0(x1))) → 01(q2(x1))
11(q1(1(x1))) → 11(q1(x1))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q5(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q0(h(x1))) → 01(q0(h(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q2(1(x1))) → 11(q2(x1))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
11(q5(h(x1))) → 01(q1(h(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
11(q0(1(x1))) → 11(q1(x1))
11(q4(x1)) → 11(x1)
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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.

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

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

01(q0(h(x1))) → 01(q0(h(x1)))
01(q0(0(x1))) → 01(q0(x1))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(x1)))

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

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

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

11(q1(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
11(q0(h(x1))) → 01(0(q1(h(x1))))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q4(x1)) → 11(x1)
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q0(0(x1))) → 01(q1(x1))
11(q0(h(x1))) → 01(0(q1(h(x1))))
11(q0(h(x1))) → 01(q1(h(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
The remaining pairs can at least be oriented weakly.

11(q1(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
11(q1(1(x1))) → 11(q1(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
11(q4(x1)) → 11(x1)
01(q1(h(x1))) → 01(0(q2(h(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(q1(x1)) = x_1   
POL(11(x1)) = (4)x_1   
POL(1(x1)) = x_1   
POL(01(x1)) = (4)x_1   
POL(q0(x1)) = 4 + x_1   
POL(q5(x1)) = x_1   
POL(q4(x1)) = x_1   
POL(h(x1)) = 0   
POL(q2(x1)) = x_1   
POL(0(x1)) = x_1   
POL(q3(x1)) = x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
h(q4(x1)) → h(0(q4(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q5(x1)) → h(0(q5(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(0(x1))) → 0(0(q0(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(0(x1))) → 1(0(q2(x1)))
0(q4(1(x1))) → 1(1(q5(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
0(q1(h(x1))) → 0(0(q2(h(x1))))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))



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

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

11(q1(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q3(x1)) → 01(x1)
01(q1(0(x1))) → 01(q2(x1))
11(q1(1(x1))) → 11(q1(x1))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q5(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 11(0(q1(h(x1))))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
11(q3(x1)) → 11(x1)
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
11(q4(x1)) → 11(x1)
01(q1(h(x1))) → 01(0(q2(h(x1))))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q3(x1)) → 01(x1)
01(q2(x1)) → 11(x1)
11(q3(x1)) → 11(x1)
11(q4(x1)) → 11(x1)
The remaining pairs can at least be oriented weakly.

11(q1(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q1(0(x1))) → 01(q2(x1))
11(q1(1(x1))) → 11(q1(x1))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q5(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 11(0(q1(h(x1))))
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(q1(x1)) = 2 + x_1   
POL(11(x1)) = (2)x_1   
POL(1(x1)) = x_1   
POL(01(x1)) = (2)x_1   
POL(q0(x1)) = 2 + x_1   
POL(q5(x1)) = 2 + x_1   
POL(q4(x1)) = 2 + x_1   
POL(h(x1)) = 0   
POL(q2(x1)) = 2 + x_1   
POL(0(x1)) = x_1   
POL(q3(x1)) = 2 + x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
h(q4(x1)) → h(0(q4(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q5(x1)) → h(0(q5(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(0(x1))) → 0(0(q0(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(0(x1))) → 1(0(q2(x1)))
0(q4(1(x1))) → 1(1(q5(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
0(q1(h(x1))) → 0(0(q2(h(x1))))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))



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

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

11(q1(0(x1))) → 01(q1(x1))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q1(0(x1))) → 01(q2(x1))
11(q1(1(x1))) → 11(q1(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q1(0(x1))) → 11(0(q1(x1)))
11(q1(h(x1))) → 11(0(q1(h(x1))))
11(q2(0(x1))) → 01(q2(x1))
11(q2(h(x1))) → 01(q2(h(x1)))
01(q1(1(x1))) → 11(q2(x1))
11(q5(0(x1))) → 01(q1(x1))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q2(1(x1))) → 11(q2(x1))
11(q2(h(x1))) → 11(0(q2(h(x1))))
11(q1(1(x1))) → 11(1(q1(x1)))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q1(h(x1))) → 01(0(q2(h(x1))))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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 12 less nodes.

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

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

11(q2(1(x1))) → 11(q2(x1))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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(q2(1(x1))) → 11(q2(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = (2)x_1   
POL(1(x1)) = 1 + (4)x_1   
POL(q2(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
                ↳ DependencyGraphProof
                  ↳ AND
                    ↳ QDP
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ QDPOrderProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ AND
                                  ↳ QDP
                                    ↳ QDPOrderProof
                                      ↳ QDP
                                        ↳ PisEmptyProof
                                  ↳ QDP
          ↳ QDP

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

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ AND
                    ↳ QDP
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ QDPOrderProof
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ AND
                                  ↳ QDP
                                  ↳ QDP
          ↳ QDP

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

11(q5(0(x1))) → 01(q1(x1))
11(q1(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q5(h(x1))) → 01(0(q1(h(x1))))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
11(q1(1(x1))) → 11(q1(x1))
11(q1(h(x1))) → 01(q1(h(x1)))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q5(1(x1))) → 11(q1(x1))
11(q5(h(x1))) → 01(q1(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q1(h(x1))) → 01(0(q2(h(x1))))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(x1)))

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

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

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

H(q3(x1)) → H(0(q3(x1)))
H(q4(x1)) → H(0(q4(x1)))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(h(x1))) → 0(0(q1(h(x1))))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(h(x1))) → 1(0(q1(h(x1))))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(h(x1))) → 0(0(q2(h(x1))))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(x1)))
1(q2(h(x1))) → 1(0(q2(h(x1))))
1(q2(1(x1))) → 1(1(q2(x1)))
0(q2(x1)) → q3(1(x1))
1(q3(x1)) → q3(1(x1))
0(q3(x1)) → q4(0(x1))
1(q4(x1)) → q4(1(x1))
0(q4(0(x1))) → 1(0(q5(x1)))
0(q4(h(x1))) → 1(0(q5(h(x1))))
0(q4(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(h(x1))) → 0(0(q1(h(x1))))
1(q5(1(x1))) → 0(1(q1(x1)))
h(q0(x1)) → h(0(q0(x1)))
h(q1(x1)) → h(0(q1(x1)))
h(q2(x1)) → h(0(q2(x1)))
h(q3(x1)) → h(0(q3(x1)))
h(q4(x1)) → h(0(q4(x1)))
h(q5(x1)) → h(0(q5(x1)))

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