NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Mixed_SRS/turing_mult.srs

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

0 QTRS
↳1 NonTerminationProof (⇒, 29 ms)
↳2 NO

(0) Obligation:

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

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

Q is empty.

(1) NonTerminationProof (COMPLETE 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