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

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

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

The set Q is empty.
We have obtained the following QTRS:

0(q0(0(x))) → q0(0(0(x)))
1(q0(0(x))) → q0(1(0(x)))
0(q0(1(x))) → q1(0(0(x)))
1(q0(1(x))) → q1(1(0(x)))
0(q1(1(x))) → q1(0(1(x)))
1(q1(1(x))) → q1(1(1(x)))
0(q1(0(x))) → q2(0(0(x)))
1(q1(0(x))) → q2(1(0(x)))
0(q2(1(x))) → q2(0(1(x)))
1(q2(1(x))) → q2(1(1(x)))
q2(0(x)) → 1(q3(x))
q3(1(x)) → 1(q3(x))
q3(0(x)) → 0(q4(x))
q4(1(x)) → 1(q4(x))
0(q4(0(x))) → q5(0(1(x)))
1(q4(0(x))) → q5(1(1(x)))
0(q5(1(x))) → q1(0(0(x)))
1(q5(1(x))) → q1(1(0(x)))
q5(0(x)) → 0(q6(x))
q6(1(x)) → 1(q6(x))
0(q7(1(x))) → q8(0(0(x)))
1(q7(1(x))) → q8(1(0(x)))
q8(0(x)) → q0(0(x))
0(q8(1(x))) → q8(0(1(x)))
1(q8(1(x))) → q8(1(1(x)))
q6(0(x)) → 0(q9(x))
0(q9(0(x))) → q7(0(1(x)))
1(q9(0(x))) → q7(1(1(x)))
q9(1(x)) → 1(q9(x))
q0(h(x)) → q0(0(h(x)))
h(q0(x)) → h(0(q0(x)))
q1(h(x)) → q1(0(h(x)))
h(q1(x)) → h(0(q1(x)))
q2(h(x)) → q2(0(h(x)))
h(q2(x)) → h(0(q2(x)))
q3(h(x)) → q3(0(h(x)))
h(q3(x)) → h(0(q3(x)))
q4(h(x)) → q4(0(h(x)))
h(q4(x)) → h(0(q4(x)))
q5(h(x)) → q5(0(h(x)))
h(q5(x)) → h(0(q5(x)))
q6(h(x)) → q6(0(h(x)))
h(q6(x)) → h(0(q6(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

0(q0(0(x))) → q0(0(0(x)))
1(q0(0(x))) → q0(1(0(x)))
0(q0(1(x))) → q1(0(0(x)))
1(q0(1(x))) → q1(1(0(x)))
0(q1(1(x))) → q1(0(1(x)))
1(q1(1(x))) → q1(1(1(x)))
0(q1(0(x))) → q2(0(0(x)))
1(q1(0(x))) → q2(1(0(x)))
0(q2(1(x))) → q2(0(1(x)))
1(q2(1(x))) → q2(1(1(x)))
q2(0(x)) → 1(q3(x))
q3(1(x)) → 1(q3(x))
q3(0(x)) → 0(q4(x))
q4(1(x)) → 1(q4(x))
0(q4(0(x))) → q5(0(1(x)))
1(q4(0(x))) → q5(1(1(x)))
0(q5(1(x))) → q1(0(0(x)))
1(q5(1(x))) → q1(1(0(x)))
q5(0(x)) → 0(q6(x))
q6(1(x)) → 1(q6(x))
0(q7(1(x))) → q8(0(0(x)))
1(q7(1(x))) → q8(1(0(x)))
q8(0(x)) → q0(0(x))
0(q8(1(x))) → q8(0(1(x)))
1(q8(1(x))) → q8(1(1(x)))
q6(0(x)) → 0(q9(x))
0(q9(0(x))) → q7(0(1(x)))
1(q9(0(x))) → q7(1(1(x)))
q9(1(x)) → 1(q9(x))
q0(h(x)) → q0(0(h(x)))
h(q0(x)) → h(0(q0(x)))
q1(h(x)) → q1(0(h(x)))
h(q1(x)) → h(0(q1(x)))
q2(h(x)) → q2(0(h(x)))
h(q2(x)) → h(0(q2(x)))
q3(h(x)) → q3(0(h(x)))
h(q3(x)) → h(0(q3(x)))
q4(h(x)) → q4(0(h(x)))
h(q4(x)) → h(0(q4(x)))
q5(h(x)) → q5(0(h(x)))
h(q5(x)) → h(0(q5(x)))
q6(h(x)) → q6(0(h(x)))
h(q6(x)) → h(0(q6(x)))

Q is empty.

We have reversed the following QTRS:
The set of rules R is

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

The set Q is empty.
We have obtained the following QTRS:

0(q0(0(x))) → q0(0(0(x)))
1(q0(0(x))) → q0(1(0(x)))
0(q0(1(x))) → q1(0(0(x)))
1(q0(1(x))) → q1(1(0(x)))
0(q1(1(x))) → q1(0(1(x)))
1(q1(1(x))) → q1(1(1(x)))
0(q1(0(x))) → q2(0(0(x)))
1(q1(0(x))) → q2(1(0(x)))
0(q2(1(x))) → q2(0(1(x)))
1(q2(1(x))) → q2(1(1(x)))
q2(0(x)) → 1(q3(x))
q3(1(x)) → 1(q3(x))
q3(0(x)) → 0(q4(x))
q4(1(x)) → 1(q4(x))
0(q4(0(x))) → q5(0(1(x)))
1(q4(0(x))) → q5(1(1(x)))
0(q5(1(x))) → q1(0(0(x)))
1(q5(1(x))) → q1(1(0(x)))
q5(0(x)) → 0(q6(x))
q6(1(x)) → 1(q6(x))
0(q7(1(x))) → q8(0(0(x)))
1(q7(1(x))) → q8(1(0(x)))
q8(0(x)) → q0(0(x))
0(q8(1(x))) → q8(0(1(x)))
1(q8(1(x))) → q8(1(1(x)))
q6(0(x)) → 0(q9(x))
0(q9(0(x))) → q7(0(1(x)))
1(q9(0(x))) → q7(1(1(x)))
q9(1(x)) → 1(q9(x))
q0(h(x)) → q0(0(h(x)))
h(q0(x)) → h(0(q0(x)))
q1(h(x)) → q1(0(h(x)))
h(q1(x)) → h(0(q1(x)))
q2(h(x)) → q2(0(h(x)))
h(q2(x)) → h(0(q2(x)))
q3(h(x)) → q3(0(h(x)))
h(q3(x)) → h(0(q3(x)))
q4(h(x)) → q4(0(h(x)))
h(q4(x)) → h(0(q4(x)))
q5(h(x)) → q5(0(h(x)))
h(q5(x)) → h(0(q5(x)))
q6(h(x)) → q6(0(h(x)))
h(q6(x)) → h(0(q6(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof

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

0(q0(0(x))) → q0(0(0(x)))
1(q0(0(x))) → q0(1(0(x)))
0(q0(1(x))) → q1(0(0(x)))
1(q0(1(x))) → q1(1(0(x)))
0(q1(1(x))) → q1(0(1(x)))
1(q1(1(x))) → q1(1(1(x)))
0(q1(0(x))) → q2(0(0(x)))
1(q1(0(x))) → q2(1(0(x)))
0(q2(1(x))) → q2(0(1(x)))
1(q2(1(x))) → q2(1(1(x)))
q2(0(x)) → 1(q3(x))
q3(1(x)) → 1(q3(x))
q3(0(x)) → 0(q4(x))
q4(1(x)) → 1(q4(x))
0(q4(0(x))) → q5(0(1(x)))
1(q4(0(x))) → q5(1(1(x)))
0(q5(1(x))) → q1(0(0(x)))
1(q5(1(x))) → q1(1(0(x)))
q5(0(x)) → 0(q6(x))
q6(1(x)) → 1(q6(x))
0(q7(1(x))) → q8(0(0(x)))
1(q7(1(x))) → q8(1(0(x)))
q8(0(x)) → q0(0(x))
0(q8(1(x))) → q8(0(1(x)))
1(q8(1(x))) → q8(1(1(x)))
q6(0(x)) → 0(q9(x))
0(q9(0(x))) → q7(0(1(x)))
1(q9(0(x))) → q7(1(1(x)))
q9(1(x)) → 1(q9(x))
q0(h(x)) → q0(0(h(x)))
h(q0(x)) → h(0(q0(x)))
q1(h(x)) → q1(0(h(x)))
h(q1(x)) → h(0(q1(x)))
q2(h(x)) → q2(0(h(x)))
h(q2(x)) → h(0(q2(x)))
q3(h(x)) → q3(0(h(x)))
h(q3(x)) → h(0(q3(x)))
q4(h(x)) → q4(0(h(x)))
h(q4(x)) → h(0(q4(x)))
q5(h(x)) → q5(0(h(x)))
h(q5(x)) → h(0(q5(x)))
q6(h(x)) → q6(0(h(x)))
h(q6(x)) → h(0(q6(x)))

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))
01(q3(x1)) → 01(x1)
11(q7(0(x1))) → 01(q8(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
Q4(h(x1)) → Q4(0(h(x1)))
01(q0(1(x1))) → 11(q0(x1))
11(q6(x1)) → Q6(1(x1))
11(q8(1(x1))) → 11(1(q8(x1)))
Q1(h(x1)) → Q1(0(h(x1)))
Q1(h(x1)) → 01(h(x1))
01(q4(0(x1))) → Q5(x1)
H(q5(x1)) → H(0(q5(x1)))
11(q5(1(x1))) → Q1(x1)
11(q0(1(x1))) → 11(q1(x1))
Q5(h(x1)) → 01(h(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
11(q6(x1)) → 11(x1)
Q0(h(x1)) → 01(h(x1))
11(q7(1(x1))) → 01(1(q8(x1)))
11(q1(0(x1))) → 01(q1(x1))
01(q5(x1)) → Q6(0(x1))
11(q0(1(x1))) → Q1(x1)
01(q5(x1)) → 01(x1)
11(q7(1(x1))) → 11(q8(x1))
11(q8(0(x1))) → 01(q8(x1))
01(q2(x1)) → Q3(1(x1))
H(q4(x1)) → 01(q4(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q2(0(x1))) → Q2(x1)
Q6(h(x1)) → 01(h(x1))
01(q0(1(x1))) → Q0(x1)
01(q4(0(x1))) → 11(0(q5(x1)))
H(q5(x1)) → 01(q5(x1))
Q0(h(x1)) → Q0(0(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q9(1(x1))) → 11(q7(x1))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
H(q6(x1)) → H(0(q6(x1)))
01(q9(0(x1))) → 11(0(q7(x1)))
11(q1(1(x1))) → Q1(x1)
11(q1(1(x1))) → 11(1(q1(x1)))
Q3(h(x1)) → Q3(0(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
Q3(h(x1)) → 01(h(x1))
01(q9(0(x1))) → 01(q7(x1))
11(q1(0(x1))) → Q1(x1)
01(q0(0(x1))) → 01(0(q0(x1)))
01(q8(x1)) → 01(q0(x1))
11(q0(0(x1))) → Q1(x1)
H(q3(x1)) → 01(q3(x1))
11(q4(x1)) → Q4(1(x1))
01(q1(0(x1))) → 01(q2(x1))
Q6(h(x1)) → Q6(0(h(x1)))
11(q1(1(x1))) → 11(q1(x1))
11(q8(0(x1))) → 11(0(q8(x1)))
Q2(h(x1)) → 01(h(x1))
11(q2(1(x1))) → Q2(x1)
11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(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(0(x1))) → Q0(x1)
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
11(q7(0(x1))) → 01(0(q8(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q8(1(x1))) → 11(q8(x1))
11(q4(x1)) → 11(x1)
01(q1(0(x1))) → Q2(x1)
11(q5(0(x1))) → Q1(x1)
01(q3(x1)) → Q4(0(x1))
01(q1(1(x1))) → Q2(x1)
01(q6(x1)) → 01(x1)
01(q9(1(x1))) → 11(1(q7(x1)))
H(q0(x1)) → 01(q0(x1))
H(q3(x1)) → H(0(q3(x1)))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q0(1(x1))) → 01(1(q0(x1)))
01(q8(x1)) → Q0(x1)
H(q4(x1)) → H(0(q4(x1)))
11(q3(x1)) → Q3(1(x1))
Q2(h(x1)) → Q2(0(h(x1)))
H(q2(x1)) → 01(q2(x1))
H(q0(x1)) → H(0(q0(x1)))
H(q1(x1)) → 01(q1(x1))
H(q6(x1)) → 01(q6(x1))
01(q1(1(x1))) → 11(q2(x1))
Q5(h(x1)) → Q5(0(h(x1)))
01(q4(1(x1))) → Q5(x1)
11(q2(1(x1))) → 11(q2(x1))
01(q4(0(x1))) → 01(q5(x1))
Q4(h(x1)) → 01(h(x1))
H(q2(x1)) → H(0(q2(x1)))
11(q9(x1)) → 11(x1)

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

11(q0(0(x1))) → 01(q1(x1))
01(q3(x1)) → 01(x1)
11(q7(0(x1))) → 01(q8(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
Q4(h(x1)) → Q4(0(h(x1)))
01(q0(1(x1))) → 11(q0(x1))
11(q6(x1)) → Q6(1(x1))
11(q8(1(x1))) → 11(1(q8(x1)))
Q1(h(x1)) → Q1(0(h(x1)))
Q1(h(x1)) → 01(h(x1))
01(q4(0(x1))) → Q5(x1)
H(q5(x1)) → H(0(q5(x1)))
11(q5(1(x1))) → Q1(x1)
11(q0(1(x1))) → 11(q1(x1))
Q5(h(x1)) → 01(h(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
11(q6(x1)) → 11(x1)
Q0(h(x1)) → 01(h(x1))
11(q7(1(x1))) → 01(1(q8(x1)))
11(q1(0(x1))) → 01(q1(x1))
01(q5(x1)) → Q6(0(x1))
11(q0(1(x1))) → Q1(x1)
01(q5(x1)) → 01(x1)
11(q7(1(x1))) → 11(q8(x1))
11(q8(0(x1))) → 01(q8(x1))
01(q2(x1)) → Q3(1(x1))
H(q4(x1)) → 01(q4(x1))
11(q5(1(x1))) → 11(q1(x1))
11(q2(0(x1))) → Q2(x1)
Q6(h(x1)) → 01(h(x1))
01(q0(1(x1))) → Q0(x1)
01(q4(0(x1))) → 11(0(q5(x1)))
H(q5(x1)) → 01(q5(x1))
Q0(h(x1)) → Q0(0(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q9(1(x1))) → 11(q7(x1))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
H(q6(x1)) → H(0(q6(x1)))
01(q9(0(x1))) → 11(0(q7(x1)))
11(q1(1(x1))) → Q1(x1)
11(q1(1(x1))) → 11(1(q1(x1)))
Q3(h(x1)) → Q3(0(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
Q3(h(x1)) → 01(h(x1))
01(q9(0(x1))) → 01(q7(x1))
11(q1(0(x1))) → Q1(x1)
01(q0(0(x1))) → 01(0(q0(x1)))
01(q8(x1)) → 01(q0(x1))
11(q0(0(x1))) → Q1(x1)
H(q3(x1)) → 01(q3(x1))
11(q4(x1)) → Q4(1(x1))
01(q1(0(x1))) → 01(q2(x1))
Q6(h(x1)) → Q6(0(h(x1)))
11(q1(1(x1))) → 11(q1(x1))
11(q8(0(x1))) → 11(0(q8(x1)))
Q2(h(x1)) → 01(h(x1))
11(q2(1(x1))) → Q2(x1)
11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(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(0(x1))) → Q0(x1)
11(q2(1(x1))) → 11(1(q2(x1)))
01(q0(0(x1))) → 01(q0(x1))
11(q7(0(x1))) → 01(0(q8(x1)))
11(q0(1(x1))) → 01(1(q1(x1)))
11(q8(1(x1))) → 11(q8(x1))
11(q4(x1)) → 11(x1)
01(q1(0(x1))) → Q2(x1)
11(q5(0(x1))) → Q1(x1)
01(q3(x1)) → Q4(0(x1))
01(q1(1(x1))) → Q2(x1)
01(q6(x1)) → 01(x1)
01(q9(1(x1))) → 11(1(q7(x1)))
H(q0(x1)) → 01(q0(x1))
H(q3(x1)) → H(0(q3(x1)))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q0(1(x1))) → 01(1(q0(x1)))
01(q8(x1)) → Q0(x1)
H(q4(x1)) → H(0(q4(x1)))
11(q3(x1)) → Q3(1(x1))
Q2(h(x1)) → Q2(0(h(x1)))
H(q2(x1)) → 01(q2(x1))
H(q0(x1)) → H(0(q0(x1)))
H(q1(x1)) → 01(q1(x1))
H(q6(x1)) → 01(q6(x1))
01(q1(1(x1))) → 11(q2(x1))
Q5(h(x1)) → Q5(0(h(x1)))
01(q4(1(x1))) → Q5(x1)
11(q2(1(x1))) → 11(q2(x1))
01(q4(0(x1))) → 01(q5(x1))
Q4(h(x1)) → 01(h(x1))
H(q2(x1)) → H(0(q2(x1)))
11(q9(x1)) → 11(x1)

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(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 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ NonTerminationProof
          ↳ QDP

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

01(q8(x1)) → 01(q0(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → Q1(x1)
11(q4(x1)) → Q4(1(x1))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
Q6(h(x1)) → Q6(0(h(x1)))
11(q8(0(x1))) → 11(0(q8(x1)))
11(q1(1(x1))) → 11(q1(x1))
Q2(h(x1)) → 01(h(x1))
11(q7(0(x1))) → 01(q8(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q2(1(x1))) → Q2(x1)
11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
Q4(h(x1)) → Q4(0(h(x1)))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q3(x1)) → 11(x1)
11(q6(x1)) → Q6(1(x1))
01(q0(1(x1))) → 11(q0(x1))
01(q0(0(x1))) → Q0(x1)
11(q8(1(x1))) → 11(1(q8(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
Q1(h(x1)) → 01(h(x1))
Q1(h(x1)) → Q1(0(h(x1)))
11(q7(0(x1))) → 01(0(q8(x1)))
01(q0(0(x1))) → 01(q0(x1))
01(q4(0(x1))) → Q5(x1)
11(q5(1(x1))) → Q1(x1)
11(q0(1(x1))) → 01(1(q1(x1)))
11(q8(1(x1))) → 11(q8(x1))
11(q4(x1)) → 11(x1)
Q5(h(x1)) → 01(h(x1))
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(0(x1))) → Q2(x1)
11(q6(x1)) → 11(x1)
11(q5(0(x1))) → Q1(x1)
01(q3(x1)) → Q4(0(x1))
01(q6(x1)) → 01(x1)
01(q1(1(x1))) → Q2(x1)
Q0(h(x1)) → 01(h(x1))
11(q7(1(x1))) → 01(1(q8(x1)))
11(q1(0(x1))) → 01(q1(x1))
01(q9(1(x1))) → 11(1(q7(x1)))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q5(x1)) → Q6(0(x1))
11(q0(1(x1))) → Q1(x1)
11(q7(1(x1))) → 11(q8(x1))
01(q5(x1)) → 01(x1)
01(q0(1(x1))) → 01(1(q0(x1)))
11(q8(0(x1))) → 01(q8(x1))
01(q8(x1)) → Q0(x1)
01(q2(x1)) → Q3(1(x1))
11(q3(x1)) → Q3(1(x1))
11(q5(1(x1))) → 11(q1(x1))
Q2(h(x1)) → Q2(0(h(x1)))
11(q2(0(x1))) → Q2(x1)
Q6(h(x1)) → 01(h(x1))
01(q0(1(x1))) → Q0(x1)
01(q4(0(x1))) → 11(0(q5(x1)))
Q0(h(x1)) → Q0(0(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
01(q9(1(x1))) → 11(q7(x1))
Q5(h(x1)) → Q5(0(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
01(q4(1(x1))) → Q5(x1)
11(q2(1(x1))) → 11(q2(x1))
11(q1(1(x1))) → Q1(x1)
Q3(h(x1)) → Q3(0(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q4(1(x1))) → 11(q5(x1))
Q3(h(x1)) → 01(h(x1))
Q4(h(x1)) → 01(h(x1))
11(q1(0(x1))) → Q1(x1)
01(q0(0(x1))) → 01(0(q0(x1)))
11(q9(x1)) → 11(x1)

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

01(q8(x1)) → 01(q0(x1))
11(q0(0(x1))) → 01(q1(x1))
11(q0(0(x1))) → Q1(x1)
11(q4(x1)) → Q4(1(x1))
01(q1(0(x1))) → 01(q2(x1))
01(q3(x1)) → 01(x1)
Q6(h(x1)) → Q6(0(h(x1)))
11(q8(0(x1))) → 11(0(q8(x1)))
11(q1(1(x1))) → 11(q1(x1))
Q2(h(x1)) → 01(h(x1))
11(q7(0(x1))) → 01(q8(x1))
11(q5(0(x1))) → 01(0(q1(x1)))
11(q2(1(x1))) → Q2(x1)
11(q1(0(x1))) → 11(0(q1(x1)))
01(q2(x1)) → 11(x1)
11(q2(0(x1))) → 01(q2(x1))
Q4(h(x1)) → Q4(0(h(x1)))
11(q2(0(x1))) → 11(0(q2(x1)))
01(q1(1(x1))) → 01(1(q2(x1)))
11(q3(x1)) → 11(x1)
11(q6(x1)) → Q6(1(x1))
01(q0(1(x1))) → 11(q0(x1))
01(q0(0(x1))) → Q0(x1)
11(q8(1(x1))) → 11(1(q8(x1)))
11(q2(1(x1))) → 11(1(q2(x1)))
Q1(h(x1)) → 01(h(x1))
Q1(h(x1)) → Q1(0(h(x1)))
11(q7(0(x1))) → 01(0(q8(x1)))
01(q0(0(x1))) → 01(q0(x1))
01(q4(0(x1))) → Q5(x1)
11(q5(1(x1))) → Q1(x1)
11(q0(1(x1))) → 01(1(q1(x1)))
11(q8(1(x1))) → 11(q8(x1))
11(q4(x1)) → 11(x1)
Q5(h(x1)) → 01(h(x1))
11(q0(1(x1))) → 11(q1(x1))
11(q0(0(x1))) → 01(0(q1(x1)))
01(q1(0(x1))) → Q2(x1)
11(q6(x1)) → 11(x1)
11(q5(0(x1))) → Q1(x1)
01(q3(x1)) → Q4(0(x1))
01(q6(x1)) → 01(x1)
01(q1(1(x1))) → Q2(x1)
Q0(h(x1)) → 01(h(x1))
11(q7(1(x1))) → 01(1(q8(x1)))
11(q1(0(x1))) → 01(q1(x1))
01(q9(1(x1))) → 11(1(q7(x1)))
01(q4(1(x1))) → 11(1(q5(x1)))
01(q5(x1)) → Q6(0(x1))
11(q0(1(x1))) → Q1(x1)
11(q7(1(x1))) → 11(q8(x1))
01(q5(x1)) → 01(x1)
01(q0(1(x1))) → 01(1(q0(x1)))
11(q8(0(x1))) → 01(q8(x1))
01(q8(x1)) → Q0(x1)
01(q2(x1)) → Q3(1(x1))
11(q3(x1)) → Q3(1(x1))
11(q5(1(x1))) → 11(q1(x1))
Q2(h(x1)) → Q2(0(h(x1)))
11(q2(0(x1))) → Q2(x1)
Q6(h(x1)) → 01(h(x1))
01(q0(1(x1))) → Q0(x1)
01(q4(0(x1))) → 11(0(q5(x1)))
Q0(h(x1)) → Q0(0(h(x1)))
11(q5(0(x1))) → 01(q1(x1))
01(q1(1(x1))) → 11(q2(x1))
01(q9(1(x1))) → 11(q7(x1))
Q5(h(x1)) → Q5(0(h(x1)))
11(q5(1(x1))) → 01(1(q1(x1)))
01(q1(0(x1))) → 01(0(q2(x1)))
01(q4(1(x1))) → Q5(x1)
11(q2(1(x1))) → 11(q2(x1))
11(q1(1(x1))) → Q1(x1)
Q3(h(x1)) → Q3(0(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q4(1(x1))) → 11(q5(x1))
Q3(h(x1)) → 01(h(x1))
Q4(h(x1)) → 01(h(x1))
11(q1(0(x1))) → Q1(x1)
01(q0(0(x1))) → 01(0(q0(x1)))
11(q9(x1)) → 11(x1)

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))


s = 01(q0(h(x1'))) evaluates to t =01(q0(h(x1')))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

01(q0(h(x1'))) → 01(q0(0(h(x1'))))
with rule q0(h(x1'')) → q0(0(h(x1''))) at position [0] and matcher [x1'' / x1']

01(q0(0(h(x1')))) → 01(q0(h(x1')))
with rule 01(q0(0(x1))) → 01(q0(x1))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.





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

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

H(q5(x1)) → H(0(q5(x1)))
H(q3(x1)) → H(0(q3(x1)))
H(q1(x1)) → H(0(q1(x1)))
H(q6(x1)) → H(0(q6(x1)))
H(q0(x1)) → H(0(q0(x1)))
H(q4(x1)) → H(0(q4(x1)))
H(q2(x1)) → H(0(q2(x1)))

The TRS R consists of the following rules:

0(q0(0(x1))) → 0(0(q0(x1)))
0(q0(1(x1))) → 0(1(q0(x1)))
1(q0(0(x1))) → 0(0(q1(x1)))
1(q0(1(x1))) → 0(1(q1(x1)))
1(q1(0(x1))) → 1(0(q1(x1)))
1(q1(1(x1))) → 1(1(q1(x1)))
0(q1(0(x1))) → 0(0(q2(x1)))
0(q1(1(x1))) → 0(1(q2(x1)))
1(q2(0(x1))) → 1(0(q2(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(1(x1))) → 1(1(q5(x1)))
1(q5(0(x1))) → 0(0(q1(x1)))
1(q5(1(x1))) → 0(1(q1(x1)))
0(q5(x1)) → q6(0(x1))
1(q6(x1)) → q6(1(x1))
1(q7(0(x1))) → 0(0(q8(x1)))
1(q7(1(x1))) → 0(1(q8(x1)))
0(q8(x1)) → 0(q0(x1))
1(q8(0(x1))) → 1(0(q8(x1)))
1(q8(1(x1))) → 1(1(q8(x1)))
0(q6(x1)) → q9(0(x1))
0(q9(0(x1))) → 1(0(q7(x1)))
0(q9(1(x1))) → 1(1(q7(x1)))
1(q9(x1)) → q9(1(x1))
h(q0(x1)) → h(0(q0(x1)))
q0(h(x1)) → q0(0(h(x1)))
h(q1(x1)) → h(0(q1(x1)))
q1(h(x1)) → q1(0(h(x1)))
h(q2(x1)) → h(0(q2(x1)))
q2(h(x1)) → q2(0(h(x1)))
h(q3(x1)) → h(0(q3(x1)))
q3(h(x1)) → q3(0(h(x1)))
h(q4(x1)) → h(0(q4(x1)))
q4(h(x1)) → q4(0(h(x1)))
h(q5(x1)) → h(0(q5(x1)))
q5(h(x1)) → q5(0(h(x1)))
h(q6(x1)) → h(0(q6(x1)))
q6(h(x1)) → q6(0(h(x1)))

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