3.29/1.56	NO
3.29/1.58	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.29/1.58	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.29/1.58	
3.29/1.58	
3.29/1.58	Termination w.r.t. Q of the given QTRS could be disproven:
3.29/1.58	
3.29/1.58	(0) QTRS
3.29/1.58	(1) NonTerminationProof [COMPLETE, 0 ms]
3.29/1.58	(2) NO
3.29/1.58	
3.29/1.58	
3.29/1.58	----------------------------------------
3.29/1.58	
3.29/1.58	(0)
3.29/1.58	Obligation:
3.29/1.58	Q restricted rewrite system:
3.29/1.58	The TRS R consists of the following rules:
3.29/1.58	
3.29/1.58	   0(q0(0(x1))) -> 0(0(q0(x1)))
3.29/1.58	   0(q0(1(x1))) -> 0(1(q0(x1)))
3.29/1.58	   1(q0(0(x1))) -> 0(0(q1(x1)))
3.29/1.58	   1(q0(1(x1))) -> 0(1(q1(x1)))
3.29/1.58	   1(q1(0(x1))) -> 1(0(q1(x1)))
3.29/1.58	   1(q1(1(x1))) -> 1(1(q1(x1)))
3.29/1.58	   0(q1(0(x1))) -> 0(0(q2(x1)))
3.29/1.58	   0(q1(1(x1))) -> 0(1(q2(x1)))
3.29/1.58	   1(q2(0(x1))) -> 1(0(q2(x1)))
3.29/1.58	   1(q2(1(x1))) -> 1(1(q2(x1)))
3.29/1.58	   0(q2(x1)) -> q3(1(x1))
3.29/1.58	   1(q3(x1)) -> q3(1(x1))
3.29/1.58	   0(q3(x1)) -> q4(0(x1))
3.29/1.58	   1(q4(x1)) -> q4(1(x1))
3.29/1.58	   0(q4(0(x1))) -> 1(0(q5(x1)))
3.29/1.58	   0(q4(1(x1))) -> 1(1(q5(x1)))
3.29/1.58	   1(q5(0(x1))) -> 0(0(q1(x1)))
3.29/1.58	   1(q5(1(x1))) -> 0(1(q1(x1)))
3.29/1.58	   0(q5(x1)) -> q6(0(x1))
3.29/1.58	   1(q6(x1)) -> q6(1(x1))
3.29/1.58	   1(q7(0(x1))) -> 0(0(q8(x1)))
3.29/1.58	   1(q7(1(x1))) -> 0(1(q8(x1)))
3.29/1.58	   0(q8(x1)) -> 0(q0(x1))
3.29/1.58	   1(q8(0(x1))) -> 1(0(q8(x1)))
3.29/1.58	   1(q8(1(x1))) -> 1(1(q8(x1)))
3.29/1.58	   0(q6(x1)) -> q9(0(x1))
3.29/1.58	   0(q9(0(x1))) -> 1(0(q7(x1)))
3.29/1.58	   0(q9(1(x1))) -> 1(1(q7(x1)))
3.29/1.58	   1(q9(x1)) -> q9(1(x1))
3.29/1.58	   h(q0(x1)) -> h(0(q0(x1)))
3.29/1.58	   q0(h(x1)) -> q0(0(h(x1)))
3.29/1.58	   h(q1(x1)) -> h(0(q1(x1)))
3.29/1.58	   q1(h(x1)) -> q1(0(h(x1)))
3.29/1.58	   h(q2(x1)) -> h(0(q2(x1)))
3.29/1.58	   q2(h(x1)) -> q2(0(h(x1)))
3.29/1.58	   h(q3(x1)) -> h(0(q3(x1)))
3.29/1.58	   q3(h(x1)) -> q3(0(h(x1)))
3.29/1.58	   h(q4(x1)) -> h(0(q4(x1)))
3.29/1.58	   q4(h(x1)) -> q4(0(h(x1)))
3.29/1.58	   h(q5(x1)) -> h(0(q5(x1)))
3.29/1.58	   q5(h(x1)) -> q5(0(h(x1)))
3.29/1.58	   h(q6(x1)) -> h(0(q6(x1)))
3.29/1.58	   q6(h(x1)) -> q6(0(h(x1)))
3.29/1.58	
3.29/1.58	Q is empty.
3.29/1.58	
3.29/1.58	----------------------------------------
3.29/1.58	
3.29/1.58	(1) NonTerminationProof (COMPLETE)
3.29/1.58	We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.
3.29/1.58	
3.29/1.58	Found the self-embedding DerivationStructure: 
3.29/1.58	"0 q0 0 h -> 0 0 q0 0 h"
3.29/1.58	0 q0 0 h -> 0 0 q0 0 h
3.29/1.58	by OverlapClosure OC 2"0 q0 0 -> 0 0 q0
3.29/1.58	by original rule (OC 1)""q0 h -> q0 0 h
3.29/1.58	by original rule (OC 1)"
3.29/1.58	
3.29/1.58	----------------------------------------
3.29/1.58	
3.29/1.58	(2)
3.29/1.58	NO
3.29/1.62	EOF