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(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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.

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

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)))

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ RRRPoloQTRSProof

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

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

Q is empty.

The following Q TRS is given: 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.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

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)))
Used ordering:
Polynomial interpretation [25]:

POL(0(x1)) = x1   
POL(1(x1)) = x1   
POL(h(x1)) = x1   
POL(q0(x1)) = 1 + x1   
POL(q1(x1)) = x1   
POL(q2(x1)) = x1   
POL(q3(x1)) = x1   
POL(q4(x1)) = x1   
POL(q5(x1)) = x1   




↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ 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(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:

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(q4(x1)) → 11(x1)
01(q1(h(x1))) → 01(0(q2(h(x1))))
11(q1(0(x1))) → 01(q1(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)))
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(q0(x1)) → H(0(q0(x1)))
H(q1(x1)) → 01(q1(x1))
H(q5(x1)) → 01(q5(x1))
01(q4(0(x1))) → 11(0(q5(x1)))
01(q1(1(x1))) → 11(q2(x1))
11(q5(0(x1))) → 01(q1(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(h(x1))) → 01(q1(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q0(0(x1))) → 01(0(q0(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(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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(q4(x1)) → 11(x1)
01(q1(h(x1))) → 01(0(q2(h(x1))))
11(q1(0(x1))) → 01(q1(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)))
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(q0(x1)) → H(0(q0(x1)))
H(q1(x1)) → 01(q1(x1))
H(q5(x1)) → 01(q5(x1))
01(q4(0(x1))) → 11(0(q5(x1)))
01(q1(1(x1))) → 11(q2(x1))
11(q5(0(x1))) → 01(q1(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(h(x1))) → 01(q1(h(x1)))
11(q1(1(x1))) → 11(1(q1(x1)))
01(q4(0(x1))) → 01(q5(x1))
01(q1(h(x1))) → 01(q2(h(x1)))
01(q4(1(x1))) → 11(q5(x1))
01(q0(0(x1))) → 01(0(q0(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(h(x1))) → 0(0(q0(h(x1))))
0(q0(1(x1))) → 0(1(q0(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 16 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ NonTerminationProof
              ↳ 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))
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)))
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(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(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(q4(x1)) → 11(x1)
01(q0(0(x1))) → 01(0(q0(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(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 used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

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))
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)))
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(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(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(q4(x1)) → 11(x1)
01(q0(0(x1))) → 01(0(q0(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(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)))


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

The DP semiunifies directly so there is only one rewrite step from 0^1(q0(h(x1))) to 0^1(q0(h(x1))).





↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ 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(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.