Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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.

(1) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure:

0 q0 0 h → 0 0 q0 0 h

0 q0 0 h → 0 0 q0 0 h
by OverlapClosure OC 2

0 q0 0 → 0 0 q0
by original rule (OC 1)

q0 h → q0 0 h
by original rule (OC 1)

(2) NO