details

property	value
status	complete
benchmark	65081.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n129.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.71948 seconds
cpu usage	10.9689
user time	10.5687
system time	0.400283
max virtual memory	3.7634896E7
max residence set size	1160688.0

stage attributes

key	value
starexec-result	YES

output

10.41/3.56 YES 10.75/3.64 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 10.75/3.64 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 10.75/3.64 10.75/3.64 10.75/3.64 Termination w.r.t. Q of the given QTRS could be proven: 10.75/3.64 10.75/3.64 (0) QTRS 10.75/3.64 (1) QTRSRRRProof [EQUIVALENT, 143 ms] 10.75/3.64 (2) QTRS 10.75/3.64 (3) DependencyPairsProof [EQUIVALENT, 5 ms] 10.75/3.64 (4) QDP 10.75/3.64 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 10.75/3.64 (6) TRUE 10.75/3.64 10.75/3.64 10.75/3.64 ---------------------------------------- 10.75/3.64 10.75/3.64 (0) 10.75/3.64 Obligation: 10.75/3.64 Q restricted rewrite system: 10.75/3.64 The TRS R consists of the following rules: 10.75/3.64 10.75/3.64 0(1(2(2(x1)))) -> 0(1(0(1(x1)))) 10.75/3.64 0(0(0(0(0(x1))))) -> 2(0(2(0(x1)))) 10.75/3.64 3(2(3(4(0(4(1(0(x1)))))))) -> 2(3(2(2(5(3(1(x1))))))) 10.75/3.64 1(4(1(0(4(1(2(2(5(3(x1)))))))))) -> 1(4(4(2(5(3(1(5(2(x1))))))))) 10.75/3.64 1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1(1(4(1(0(4(2(3(2(1(x1)))))))))) 10.75/3.64 2(5(1(2(4(5(1(3(1(5(x1)))))))))) -> 2(4(3(0(3(4(2(4(5(x1))))))))) 10.75/3.64 2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) -> 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) 10.75/3.64 5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) -> 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) 10.75/3.64 1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) -> 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) 10.75/3.64 1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) -> 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) 10.75/3.64 4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) -> 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) 10.75/3.64 5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) -> 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) 10.75/3.64 0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) -> 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) 10.75/3.64 4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) -> 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) 10.75/3.64 3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) -> 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) 10.75/3.64 4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) -> 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) 10.75/3.64 4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) -> 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) 10.75/3.64 4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) -> 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) 10.75/3.64 10.75/3.64 Q is empty. 10.75/3.64 10.75/3.64 ---------------------------------------- 10.75/3.64 10.75/3.64 (1) QTRSRRRProof (EQUIVALENT) 10.75/3.64 Used ordering: 10.75/3.64 Polynomial interpretation [POLO]: 10.75/3.64 10.75/3.64 POL(0(x_1)) = 13 + x_1 10.75/3.64 POL(1(x_1)) = 9 + x_1 10.75/3.64 POL(2(x_1)) = 14 + x_1 10.75/3.64 POL(3(x_1)) = 11 + x_1 10.75/3.64 POL(4(x_1)) = 8 + x_1 10.75/3.64 POL(5(x_1)) = 10 + x_1 10.75/3.64 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 10.75/3.64 10.75/3.64 0(1(2(2(x1)))) -> 0(1(0(1(x1)))) 10.75/3.64 0(0(0(0(0(x1))))) -> 2(0(2(0(x1)))) 10.75/3.64 3(2(3(4(0(4(1(0(x1)))))))) -> 2(3(2(2(5(3(1(x1))))))) 10.75/3.64 1(4(1(0(4(1(2(2(5(3(x1)))))))))) -> 1(4(4(2(5(3(1(5(2(x1))))))))) 10.75/3.64 2(5(1(2(4(5(1(3(1(5(x1)))))))))) -> 2(4(3(0(3(4(2(4(5(x1))))))))) 10.75/3.64 2(3(2(4(3(2(3(4(4(0(0(0(x1)))))))))))) -> 2(4(2(0(2(2(2(2(4(3(0(x1))))))))))) 10.75/3.64 5(0(2(4(2(4(1(4(4(5(1(4(x1)))))))))))) -> 1(4(5(3(3(2(3(2(3(3(4(x1))))))))))) 10.75/3.64 1(0(1(3(5(5(1(2(5(2(3(5(4(x1))))))))))))) -> 1(0(3(4(1(2(3(4(5(3(3(5(4(x1))))))))))))) 10.75/3.64 1(4(4(2(5(3(1(5(1(2(1(5(0(x1))))))))))))) -> 3(5(2(5(4(1(5(2(4(1(3(2(x1)))))))))))) 10.75/3.64 4(1(2(5(1(1(0(0(5(4(1(3(1(x1))))))))))))) -> 4(1(1(2(2(3(5(1(4(2(3(1(x1)))))))))))) 10.75/3.64 5(5(3(2(0(3(4(0(0(3(1(4(3(x1))))))))))))) -> 5(5(3(5(0(3(1(5(2(3(1(3(4(x1))))))))))))) 10.75/3.64 0(5(3(2(4(0(2(1(2(3(3(4(3(3(x1)))))))))))))) -> 0(5(4(4(5(3(4(0(1(0(1(3(1(2(3(x1))))))))))))))) 10.75/3.64 4(3(0(5(5(2(5(2(3(5(3(0(2(2(4(x1))))))))))))))) -> 4(3(3(4(3(1(4(1(5(0(0(5(1(5(3(4(x1)))))))))))))))) 10.75/3.64 3(1(5(4(1(2(0(0(1(0(0(0(2(0(4(5(x1)))))))))))))))) -> 3(0(2(2(3(3(1(3(2(2(1(2(2(5(5(x1))))))))))))))) 10.75/3.64 4(1(1(0(5(4(2(0(4(0(5(1(2(0(3(1(x1)))))))))))))))) -> 2(4(3(1(4(1(4(0(1(1(0(5(4(0(5(0(x1)))))))))))))))) 10.75/3.64 4(0(2(2(4(4(1(1(1(1(0(4(1(5(1(2(0(1(x1)))))))))))))))))) -> 4(5(5(2(5(0(2(1(5(2(4(1(1(1(5(3(2(x1))))))))))))))))) 10.75/3.64 4(4(4(4(3(1(1(3(3(4(2(2(4(0(3(5(4(2(5(2(3(x1))))))))))))))))))))) -> 2(5(5(4(0(1(5(3(3(5(0(1(5(1(5(4(2(4(2(2(x1)))))))))))))))))))) 10.75/3.64 10.75/3.64 10.75/3.64 10.75/3.64 10.75/3.64 ---------------------------------------- 10.75/3.64 10.75/3.64 (2) 10.75/3.64 Obligation: 10.75/3.64 Q restricted rewrite system: 10.75/3.64 The TRS R consists of the following rules: 10.75/3.64 10.75/3.64 1(4(2(3(1(4(0(2(1(1(x1)))))))))) -> 1(1(4(1(0(4(2(3(2(1(x1)))))))))) 10.75/3.64 10.75/3.64 Q is empty. 10.75/3.64 10.75/3.64 ---------------------------------------- 10.75/3.64 10.75/3.64 (3) DependencyPairsProof (EQUIVALENT) 10.75/3.64 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 10.75/3.64 ---------------------------------------- 10.75/3.64 10.75/3.64 (4) 10.75/3.64 Obligation: 10.75/3.64 Q DP problem: 10.75/3.64 The TRS P consists of the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS_Standard 2019-03-29 03.29