Termination w.r.t. Q proof of /tmp/tmpt9lqrG/turing

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(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

  ↳ 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.

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

  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

Q6(h(x1)) → Q6(0(h(x1)))
Q4(h(x1)) → Q4(0(h(x1)))
Q6(h(x1)) → 0¹(h(x1))
Q4(h(x1)) → 0¹(h(x1))

The remaining pairs can at least be oriented weakly.

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)
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))
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)
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))
1¹(q1(0(x1))) → Q1(x1)
0¹(q0(0(x1))) → 0¹(0(q0(x1)))
1¹(q9(x1)) → 1¹(x1)

Used ordering: Polynomial interpretation [25,35]:

POL(Q1(x₁)) = 0
POL(1¹(x₁)) = 0
POL(0¹(x₁)) = 0
POL(q8(x₁)) = 4
POL(q0(x₁)) = 1/4 + (5/2)x_1
POL(h(x₁)) = 3/4 + (4)x_1
POL(Q6(x₁)) = (1/4)x_1
POL(q6(x₁)) = (13/4)x_1
POL(Q4(x₁)) = (1/4)x_1
POL(q2(x₁)) = 4
POL(Q0(x₁)) = 0
POL(Q2(x₁)) = 0
POL(q3(x₁)) = (4)x_1
POL(q9(x₁)) = 0
POL(q1(x₁)) = 7/2
POL(Q5(x₁)) = 0
POL(1(x₁)) = 0
POL(q7(x₁)) = 2 + (13/4)x_1
POL(q5(x₁)) = 2
POL(Q3(x₁)) = 0
POL(q4(x₁)) = 0
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ 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)
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))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q3(x1)) → 1¹(x1)
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q6(x1)) → Q6(1(x1))
0¹(q0(0(x1))) → Q0(x1)
Q1(h(x1)) → Q1(0(h(x1)))
Q1(h(x1)) → 0¹(h(x1))
1¹(q2(1(x1))) → 1¹(1(q2(x1)))
1¹(q8(1(x1))) → 1¹(1(q8(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)
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q5(x1)) → 0¹(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q8(x1)) → Q0(x1)
1¹(q8(0(x1))) → 0¹(q8(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)
0¹(q0(1(x1))) → Q0(x1)
0¹(q4(0(x1))) → 1¹(0(q5(x1)))
Q0(h(x1)) → Q0(0(h(x1)))
0¹(q1(1(x1))) → 1¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
Q5(h(x1)) → Q5(0(h(x1)))
0¹(q9(1(x1))) → 1¹(q7(x1))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q4(1(x1))) → Q5(x1)
1¹(q1(1(x1))) → Q1(x1)
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
Q3(h(x1)) → Q3(0(h(x1)))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
Q3(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → Q1(x1)
0¹(q1(0(x1))) → 0¹(q2(x1))
0¹(q3(x1)) → 0¹(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))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q3(x1)) → 1¹(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(q1(1(x1))) → Q2(x1)
Q0(h(x1)) → 0¹(h(x1))
0¹(q6(x1)) → 0¹(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)))
1¹(q0(1(x1))) → Q1(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q5(x1)) → 0¹(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)
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))
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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

Q3(h(x1)) → Q3(0(h(x1)))
Q3(h(x1)) → 0¹(h(x1))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(Q1(x₁)) = 0
POL(1¹(x₁)) = 0
POL(0¹(x₁)) = 0
POL(q8(x₁)) = 0
POL(q0(x₁)) = 0
POL(h(x₁)) = 4
POL(q6(x₁)) = (4)x_1
POL(q2(x₁)) = 0
POL(Q0(x₁)) = 0
POL(Q2(x₁)) = 0
POL(q9(x₁)) = (3/2)x_1
POL(q3(x₁)) = 0
POL(q1(x₁)) = 0
POL(Q5(x₁)) = 0
POL(1(x₁)) = 0
POL(q7(x₁)) = 4 + (1/2)x_1
POL(q5(x₁)) = 0
POL(Q3(x₁)) = (4)x_1
POL(q4(x₁)) = (5/4)x_1
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

          ↳ QDP

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

1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → Q1(x1)
0¹(q1(0(x1))) → 0¹(q2(x1))
0¹(q3(x1)) → 0¹(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))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q3(x1)) → 1¹(x1)
0¹(q0(1(x1))) → 1¹(q0(x1))
0¹(q0(0(x1))) → Q0(x1)
Q1(h(x1)) → Q1(0(h(x1)))
Q1(h(x1)) → 0¹(h(x1))
1¹(q2(1(x1))) → 1¹(1(q2(x1)))
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
1¹(q7(0(x1))) → 0¹(0(q8(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(q1(1(x1))) → Q2(x1)
Q0(h(x1)) → 0¹(h(x1))
0¹(q6(x1)) → 0¹(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)))
1¹(q0(1(x1))) → Q1(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q5(x1)) → 0¹(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)
0¹(q0(1(x1))) → Q0(x1)
0¹(q4(0(x1))) → 1¹(0(q5(x1)))
Q0(h(x1)) → Q0(0(h(x1)))
0¹(q1(1(x1))) → 1¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
Q5(h(x1)) → Q5(0(h(x1)))
0¹(q9(1(x1))) → 1¹(q7(x1))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q4(1(x1))) → Q5(x1)
1¹(q1(1(x1))) → Q1(x1)
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

          ↳ QDP

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

1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → Q1(x1)
0¹(q3(x1)) → 0¹(x1)
0¹(q1(0(x1))) → 0¹(q2(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))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q3(x1)) → 1¹(x1)
0¹(q0(1(x1))) → 1¹(q0(x1))
0¹(q0(0(x1))) → Q0(x1)
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
Q1(h(x1)) → 0¹(h(x1))
Q1(h(x1)) → Q1(0(h(x1)))
1¹(q2(1(x1))) → 1¹(1(q2(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(q1(1(x1))) → Q2(x1)
Q0(h(x1)) → 0¹(h(x1))
0¹(q6(x1)) → 0¹(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)))
1¹(q0(1(x1))) → Q1(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q5(x1)) → 0¹(x1)
1¹(q8(0(x1))) → 0¹(q8(x1))
0¹(q8(x1)) → Q0(x1)
1¹(q5(1(x1))) → 1¹(q1(x1))
Q2(h(x1)) → Q2(0(h(x1)))
1¹(q2(0(x1))) → Q2(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)
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → Q1(x1)
0¹(q1(0(x1))) → 0¹(q2(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(q1(0(x1))) → 1¹(0(q1(x1)))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
Q1(h(x1)) → Q1(0(h(x1)))
1¹(q2(1(x1))) → 1¹(1(q2(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(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(0(x1))) → Q1(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)))
1¹(q0(1(x1))) → Q1(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
Q2(h(x1)) → Q2(0(h(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)
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q1(0(x1))) → Q1(x1)
0¹(q0(0(x1))) → 0¹(0(q0(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(Q1(x₁)) = 1
POL(1¹(x₁)) = (1/4)x_1
POL(0¹(x₁)) = (1/4)x_1
POL(q8(x₁)) = 4 + x_1
POL(q0(x₁)) = 4 + x_1
POL(h(x₁)) = 0
POL(q6(x₁)) = 4 + x_1
POL(q2(x₁)) = 4 + x_1
POL(Q0(x₁)) = 1/4 + (1/4)x_1
POL(Q2(x₁)) = (1/4)x_1
POL(q9(x₁)) = 4 + x_1
POL(q3(x₁)) = 4 + x_1
POL(q1(x₁)) = 4 + x_1
POL(Q5(x₁)) = 1 + (1/4)x_1
POL(1(x₁)) = x_1
POL(q7(x₁)) = 4 + x_1
POL(q5(x₁)) = 4 + x_1
POL(q4(x₁)) = 4 + x_1
POL(0(x₁)) = x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

          ↳ QDP

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

1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → Q1(x1)
0¹(q1(0(x1))) → 0¹(q2(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(q1(0(x1))) → 1¹(0(q1(x1)))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(1(q2(x1)))
Q1(h(x1)) → Q1(0(h(x1)))
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
0¹(q4(0(x1))) → Q5(x1)
0¹(q0(0(x1))) → 0¹(q0(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
1¹(q5(1(x1))) → Q1(x1)
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(0(x1))) → Q1(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)))
1¹(q0(1(x1))) → Q1(x1)
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
Q2(h(x1)) → Q2(0(h(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)
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q1(0(x1))) → Q1(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(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 5 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

          ↳ QDP

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

Q5(h(x1)) → Q5(0(h(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

Q5(h(x1)) → Q5(0(h(x1)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(q8(x₁)) = (4)x_1
POL(q0(x₁)) = x_1
POL(h(x₁)) = 4
POL(q6(x₁)) = (1/4)x_1
POL(q2(x₁)) = 9/4 + (4)x_1
POL(q3(x₁)) = (3/2)x_1
POL(q9(x₁)) = 0
POL(Q5(x₁)) = (4)x_1
POL(q1(x₁)) = 11/4 + (13/4)x_1
POL(1(x₁)) = 0
POL(q7(x₁)) = 4 + (7/2)x_1
POL(q5(x₁)) = 4
POL(q4(x₁)) = 0
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ PisEmptyProof

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ 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(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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

          ↳ QDP

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

Q0(h(x1)) → Q0(0(h(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

Q0(h(x1)) → Q0(0(h(x1)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(q8(x₁)) = 0
POL(q0(x₁)) = (9/4)x_1
POL(h(x₁)) = 1/2 + (5/4)x_1
POL(q6(x₁)) = (11/4)x_1
POL(q2(x₁)) = 1/4 + (4)x_1
POL(Q0(x₁)) = (1/4)x_1
POL(q3(x₁)) = 0
POL(q9(x₁)) = x_1
POL(q1(x₁)) = 0
POL(1(x₁)) = 0
POL(q7(x₁)) = 9/4 + (13/4)x_1
POL(q5(x₁)) = (4)x_1
POL(q4(x₁)) = 0
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ PisEmptyProof

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ 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(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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                    ↳ QDP

                                    ↳ QDP

          ↳ QDP

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

Q1(h(x1)) → Q1(0(h(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

Q1(h(x1)) → Q1(0(h(x1)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(Q1(x₁)) = x_1
POL(q8(x₁)) = 1/2
POL(q0(x₁)) = 1/4 + (4)x_1
POL(h(x₁)) = 2
POL(q6(x₁)) = (4)x_1
POL(q2(x₁)) = 3/4 + (3/4)x_1
POL(q3(x₁)) = 0
POL(q9(x₁)) = 0
POL(q1(x₁)) = 4 + (7/2)x_1
POL(1(x₁)) = 0
POL(q7(x₁)) = 0
POL(q5(x₁)) = (7/2)x_1
POL(q4(x₁)) = (7/2)x_1
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ PisEmptyProof

                                    ↳ QDP

                                    ↳ 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(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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                    ↳ QDP

          ↳ QDP

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

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(1(q7(x1)))
0¹(q4(1(x1))) → 1¹(1(q5(x1)))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q1(0(x1))) → 0¹(q2(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q8(0(x1))) → 1¹(0(q8(x1)))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q7(0(x1))) → 0¹(q8(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(q2(0(x1))) → 0¹(q2(x1))
0¹(q4(0(x1))) → 1¹(0(q5(x1)))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q1(1(x1))) → 1¹(q2(x1))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
1¹(q2(1(x1))) → 1¹(1(q2(x1)))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(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(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

0¹(q9(1(x1))) → 1¹(1(q7(x1)))
0¹(q4(1(x1))) → 1¹(1(q5(x1)))
1¹(q8(0(x1))) → 1¹(0(q8(x1)))
1¹(q1(0(x1))) → 1¹(0(q1(x1)))
0¹(q4(0(x1))) → 1¹(0(q5(x1)))
1¹(q2(0(x1))) → 1¹(0(q2(x1)))
1¹(q1(1(x1))) → 1¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(1(q8(x1)))
1¹(q2(1(x1))) → 1¹(1(q2(x1)))

The remaining pairs can at least be oriented weakly.

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
0¹(q1(0(x1))) → 0¹(q2(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q1(1(x1))) → 1¹(q2(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
0¹(q0(0(x1))) → 0¹(0(q0(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = (2)x_1
POL(0¹(x₁)) = 2
POL(q8(x₁)) = 1
POL(q0(x₁)) = 1
POL(h(x₁)) = (4)x_1
POL(q6(x₁)) = 0
POL(q2(x₁)) = 1
POL(q3(x₁)) = 0
POL(q9(x₁)) = 0
POL(q1(x₁)) = 1
POL(1(x₁)) = 0
POL(q7(x₁)) = 1
POL(q5(x₁)) = 1
POL(q4(x₁)) = 0
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                    ↳ QDP

          ↳ QDP

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

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q1(0(x1))) → 0¹(q2(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q1(1(x1))) → 1¹(q2(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(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(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

0¹(q1(0(x1))) → 0¹(q2(x1))
0¹(q1(1(x1))) → 1¹(q2(x1))

The remaining pairs can at least be oriented weakly.

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q7(1(x1))) → 1¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
0¹(q0(0(x1))) → 0¹(0(q0(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = 4 + (1/2)x_1
POL(0¹(x₁)) = 4 + (1/2)x_1
POL(q8(x₁)) = 1/2
POL(q0(x₁)) = 1/2
POL(h(x₁)) = 9/4
POL(q6(x₁)) = 1/2
POL(q2(x₁)) = 0
POL(q3(x₁)) = 1/4 + (1/2)x_1
POL(q9(x₁)) = 1/2
POL(q1(x₁)) = 1/2
POL(1(x₁)) = 1/2
POL(q7(x₁)) = 1/2
POL(q5(x₁)) = 1/2
POL(q4(x₁)) = 1/2
POL(0(x₁)) = 1/2

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                    ↳ QDP

          ↳ QDP

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

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q2(0(x1))) → 0¹(q2(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(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(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

1¹(q2(0(x1))) → 0¹(q2(x1))

The remaining pairs can at least be oriented weakly.

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
0¹(q0(0(x1))) → 0¹(0(q0(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = (1/2)x_1
POL(0¹(x₁)) = 0
POL(q8(x₁)) = 0
POL(q0(x₁)) = 0
POL(h(x₁)) = 3/2
POL(q6(x₁)) = (4)x_1
POL(q2(x₁)) = 1/4
POL(q3(x₁)) = 0
POL(q9(x₁)) = (4)x_1
POL(q1(x₁)) = 0
POL(1(x₁)) = 0
POL(q7(x₁)) = 0
POL(q5(x₁)) = 0
POL(q4(x₁)) = 0
POL(0(x₁)) = 0

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

                                    ↳ QDP

          ↳ QDP

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

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q7(1(x1))) → 1¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(0(x1))) → 0¹(q5(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(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(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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

0¹(q4(0(x1))) → 0¹(q5(x1))

The remaining pairs can at least be oriented weakly.

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q7(1(x1))) → 1¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(x1)))
0¹(q0(0(x1))) → 0¹(0(q0(x1)))

Used ordering: Polynomial interpretation [25,35]:

POL(1¹(x₁)) = 1/2
POL(0¹(x₁)) = (1/4)x_1
POL(q8(x₁)) = 2
POL(q0(x₁)) = 2
POL(h(x₁)) = 15/4
POL(q6(x₁)) = 2
POL(q2(x₁)) = 0
POL(q3(x₁)) = 2
POL(q9(x₁)) = 1/2 + (3/4)x_1
POL(q1(x₁)) = 2
POL(1(x₁)) = 2
POL(q7(x₁)) = 4
POL(q5(x₁)) = 0
POL(q4(x₁)) = 2
POL(0(x₁)) = 2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                      ↳ QDPOrderProof

                                        ↳ QDP

                                          ↳ QDPOrderProof

                                            ↳ QDP

                                              ↳ QDPOrderProof

                                                ↳ QDP

                                                  ↳ QDPOrderProof

                                                    ↳ QDP

                                    ↳ QDP

          ↳ QDP

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

1¹(q7(1(x1))) → 0¹(1(q8(x1)))
1¹(q1(0(x1))) → 0¹(q1(x1))
0¹(q8(x1)) → 0¹(q0(x1))
1¹(q0(0(x1))) → 0¹(q1(x1))
1¹(q7(1(x1))) → 1¹(q8(x1))
0¹(q0(1(x1))) → 0¹(1(q0(x1)))
1¹(q8(0(x1))) → 0¹(q8(x1))
1¹(q1(1(x1))) → 1¹(q1(x1))
1¹(q7(0(x1))) → 0¹(q8(x1))
1¹(q5(1(x1))) → 1¹(q1(x1))
1¹(q5(0(x1))) → 0¹(0(q1(x1)))
1¹(q5(0(x1))) → 0¹(q1(x1))
0¹(q9(1(x1))) → 1¹(q7(x1))
0¹(q1(1(x1))) → 0¹(1(q2(x1)))
1¹(q5(1(x1))) → 0¹(1(q1(x1)))
0¹(q1(0(x1))) → 0¹(0(q2(x1)))
0¹(q0(1(x1))) → 1¹(q0(x1))
1¹(q2(1(x1))) → 1¹(q2(x1))
1¹(q7(0(x1))) → 0¹(0(q8(x1)))
0¹(q0(0(x1))) → 0¹(q0(x1))
0¹(q4(1(x1))) → 1¹(q5(x1))
1¹(q0(1(x1))) → 0¹(1(q1(x1)))
1¹(q8(1(x1))) → 1¹(q8(x1))
1¹(q0(1(x1))) → 1¹(q1(x1))
1¹(q0(0(x1))) → 0¹(0(q1(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(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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ AND

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

                                    ↳ QDP

          ↳ QDP

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

Q2(h(x1)) → Q2(0(h(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

  ↳ 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.