3.10/1.53	NO
3.10/1.55	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.10/1.55	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.10/1.55	
3.10/1.55	
3.10/1.55	Termination w.r.t. Q of the given QTRS could be disproven:
3.10/1.55	
3.10/1.55	(0) QTRS
3.10/1.55	(1) NonTerminationProof [COMPLETE, 0 ms]
3.10/1.55	(2) NO
3.10/1.55	
3.10/1.55	
3.10/1.55	----------------------------------------
3.10/1.55	
3.10/1.55	(0)
3.10/1.55	Obligation:
3.10/1.55	Q restricted rewrite system:
3.10/1.55	The TRS R consists of the following rules:
3.10/1.55	
3.10/1.55	   0(q0(0(x1))) -> 0(0(q0(x1)))
3.10/1.55	   0(q0(h(x1))) -> 0(0(q0(h(x1))))
3.10/1.55	   0(q0(1(x1))) -> 0(1(q0(x1)))
3.10/1.55	   1(q0(0(x1))) -> 0(0(q1(x1)))
3.10/1.55	   1(q0(h(x1))) -> 0(0(q1(h(x1))))
3.10/1.55	   1(q0(1(x1))) -> 0(1(q1(x1)))
3.10/1.55	   1(q1(0(x1))) -> 1(0(q1(x1)))
3.10/1.55	   1(q1(h(x1))) -> 1(0(q1(h(x1))))
3.10/1.55	   1(q1(1(x1))) -> 1(1(q1(x1)))
3.10/1.55	   0(q1(0(x1))) -> 0(0(q2(x1)))
3.10/1.55	   0(q1(h(x1))) -> 0(0(q2(h(x1))))
3.10/1.55	   0(q1(1(x1))) -> 0(1(q2(x1)))
3.10/1.55	   1(q2(0(x1))) -> 1(0(q2(x1)))
3.10/1.55	   1(q2(h(x1))) -> 1(0(q2(h(x1))))
3.10/1.55	   1(q2(1(x1))) -> 1(1(q2(x1)))
3.10/1.55	   0(q2(x1)) -> q3(1(x1))
3.10/1.55	   1(q3(x1)) -> q3(1(x1))
3.10/1.55	   0(q3(x1)) -> q4(0(x1))
3.10/1.55	   1(q4(x1)) -> q4(1(x1))
3.10/1.55	   0(q4(0(x1))) -> 1(0(q5(x1)))
3.10/1.55	   0(q4(h(x1))) -> 1(0(q5(h(x1))))
3.10/1.55	   0(q4(1(x1))) -> 1(1(q5(x1)))
3.10/1.55	   1(q5(0(x1))) -> 0(0(q1(x1)))
3.10/1.55	   1(q5(h(x1))) -> 0(0(q1(h(x1))))
3.10/1.55	   1(q5(1(x1))) -> 0(1(q1(x1)))
3.10/1.55	   h(q0(x1)) -> h(0(q0(x1)))
3.10/1.55	   h(q1(x1)) -> h(0(q1(x1)))
3.10/1.55	   h(q2(x1)) -> h(0(q2(x1)))
3.10/1.55	   h(q3(x1)) -> h(0(q3(x1)))
3.10/1.55	   h(q4(x1)) -> h(0(q4(x1)))
3.10/1.55	   h(q5(x1)) -> h(0(q5(x1)))
3.10/1.55	
3.10/1.55	Q is empty.
3.10/1.55	
3.10/1.55	----------------------------------------
3.10/1.55	
3.10/1.55	(1) NonTerminationProof (COMPLETE)
3.10/1.55	We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.
3.10/1.55	
3.10/1.55	Found the self-embedding DerivationStructure: 
3.10/1.55	"0 q0 h -> 0 0 q0 h"
3.10/1.55	0 q0 h -> 0 0 q0 h
3.10/1.55	by original rule (OC 1)
3.10/1.55	
3.10/1.55	----------------------------------------
3.10/1.55	
3.10/1.55	(2)
3.10/1.55	NO
3.10/1.58	EOF