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(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.

(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 h → 0 0 q0 h

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

(2) NO