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

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

0 QTRS
↳1 NonTerminationProof (⇒, 0 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(h(x))) → 0(0(q0(h(x))))
0(q0(1(x))) → 0(1(q0(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q0(h(x))) → 0(0(q1(h(x))))
1(q0(1(x))) → 0(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
1(q1(h(x))) → 1(0(q1(h(x))))
1(q1(1(x))) → 1(1(q1(x)))
0(q1(0(x))) → 0(0(q2(x)))
0(q1(h(x))) → 0(0(q2(h(x))))
0(q1(1(x))) → 0(1(q2(x)))
1(q2(0(x))) → 1(0(q2(x)))
1(q2(h(x))) → 1(0(q2(h(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(h(x))) → 1(0(q5(h(x))))
0(q4(1(x))) → 1(1(q5(x)))
1(q5(0(x))) → 0(0(q1(x)))
1(q5(h(x))) → 0(0(q1(h(x))))
1(q5(1(x))) → 0(1(q1(x)))
h(q0(x)) → h(0(q0(x)))
h(q1(x)) → h(0(q1(x)))
h(q2(x)) → h(0(q2(x)))
h(q3(x)) → h(0(q3(x)))
h(q4(x)) → h(0(q4(x)))
h(q5(x)) → h(0(q5(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 h → 0 0 q0 h

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

(2) NO