SRS_Standard 2019-03-29 03.29 pair #432289309

details

property	value
status	complete
benchmark	turing_copy.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n114.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.58575 seconds
cpu usage	3.28741
user time	3.12773
system time	0.159684
max virtual memory	1.8609136E7
max residence set size	225832.0

stage attributes

key	value
starexec-result	NO

output

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3.10/1.55	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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

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