details

property	value
status	complete
benchmark	230819.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.59246 seconds
cpu usage	18.6888
user time	17.3746
system time	1.3142
max virtual memory	3.8312736E7
max residence set size	3363660.0

stage attributes

key	value
starexec-result	YES

output

11.53/3.75 YES 18.29/5.51 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 18.29/5.51 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 18.29/5.51 18.29/5.51 18.29/5.51 Termination w.r.t. Q of the given QTRS could be proven: 18.29/5.51 18.29/5.51 (0) QTRS 18.29/5.51 (1) QTRSRRRProof [EQUIVALENT, 93 ms] 18.29/5.51 (2) QTRS 18.29/5.51 (3) Overlay + Local Confluence [EQUIVALENT, 3 ms] 18.29/5.51 (4) QTRS 18.29/5.51 (5) DependencyPairsProof [EQUIVALENT, 26 ms] 18.29/5.51 (6) QDP 18.29/5.51 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 18.29/5.51 (8) TRUE 18.29/5.51 18.29/5.51 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (0) 18.29/5.51 Obligation: 18.29/5.51 Q restricted rewrite system: 18.29/5.51 The TRS R consists of the following rules: 18.29/5.51 18.29/5.51 0(x1) -> 1(x1) 18.29/5.51 0(0(x1)) -> 0(x1) 18.29/5.51 3(4(5(x1))) -> 4(3(5(x1))) 18.29/5.51 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 0(1(0(1(0(0(1(0(0(0(0(0(0(1(0(1(1(0(1(1(1(0(1(0(0(1(0(1(1(0(0(0(1(1(1(1(1(1(0(1(0(0(0(1(1(0(0(0(1(0(1(1(1(0(1(1(0(1(1(0(0(1(1(1(1(0(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(0(0(1(1(1(1(0(0(0(1(1(0(0(0(0(1(1(0(0(0(1(0(1(0(0(1(0(1(0(1(0(0(0(1(1(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(1(1(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 18.29/5.51 0(0(1(1(0(0(1(1(1(0(0(0(1(1(1(0(1(0(1(1(1(1(0(0(0(1(0(1(1(0(0(1(1(0(0(0(1(1(1(0(1(0(1(0(0(1(0(0(0(1(1(0(1(0(0(1(0(0(1(0(0(1(0(0(0(0(1(0(0(1(0(1(1(0(0(1(1(0(1(0(1(0(1(1(1(0(0(0(1(0(1(0(0(1(0(1(0(1(0(0(0(0(0(0(1(1(0(1(1(1(0(1(0(1(1(0(1(0(0(0(1(1(1(1(0(1(1(1(1(1(0(1(0(1(1(0(1(1(0(1(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 18.29/5.51 18.29/5.51 Q is empty. 18.29/5.51 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (1) QTRSRRRProof (EQUIVALENT) 18.29/5.51 Used ordering: 18.29/5.51 Polynomial interpretation [POLO]: 18.29/5.51 18.29/5.51 POL(0(x_1)) = 10 + x_1 18.29/5.51 POL(1(x_1)) = 9 + x_1 18.29/5.51 POL(2(x_1)) = 14 + x_1 18.29/5.51 POL(3(x_1)) = x_1 18.29/5.51 POL(4(x_1)) = x_1 18.29/5.51 POL(5(x_1)) = x_1 18.29/5.51 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 18.29/5.51 18.29/5.51 0(x1) -> 1(x1) 18.29/5.51 0(0(x1)) -> 0(x1) 18.29/5.51 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 0(1(0(1(0(0(1(0(0(0(0(0(0(1(0(1(1(0(1(1(1(0(1(0(0(1(0(1(1(0(0(0(1(1(1(1(1(1(0(1(0(0(0(1(1(0(0(0(1(0(1(1(1(0(1(1(0(1(1(0(0(1(1(1(1(0(1(1(1(1(1(0(0(1(0(0(1(1(1(1(0(1(1(0(0(0(1(1(1(1(0(0(0(1(1(0(0(0(0(1(1(0(0(0(1(0(1(0(0(1(0(1(0(1(0(0(0(1(1(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(1(1(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 18.29/5.51 0(0(1(1(0(0(1(1(1(0(0(0(1(1(1(0(1(0(1(1(1(1(0(0(0(1(0(1(1(0(0(1(1(0(0(0(1(1(1(0(1(0(1(0(0(1(0(0(0(1(1(0(1(0(0(1(0(0(1(0(0(1(0(0(0(0(1(0(0(1(0(1(1(0(0(1(1(0(1(0(1(0(1(1(1(0(0(0(1(0(1(0(0(1(0(1(0(1(0(0(0(0(0(0(1(1(0(1(1(1(0(1(0(1(1(0(1(0(0(0(1(1(1(1(0(1(1(1(1(1(0(1(0(1(1(0(1(1(0(1(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(x1))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 18.29/5.51 18.29/5.51 18.29/5.51 18.29/5.51 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (2) 18.29/5.51 Obligation: 18.29/5.51 Q restricted rewrite system: 18.29/5.51 The TRS R consists of the following rules: 18.29/5.51 18.29/5.51 3(4(5(x1))) -> 4(3(5(x1))) 18.29/5.51 18.29/5.51 Q is empty. 18.29/5.51 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (3) Overlay + Local Confluence (EQUIVALENT) 18.29/5.51 The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (4) 18.29/5.51 Obligation: 18.29/5.51 Q restricted rewrite system: 18.29/5.51 The TRS R consists of the following rules: 18.29/5.51 18.29/5.51 3(4(5(x1))) -> 4(3(5(x1))) 18.29/5.51 18.29/5.51 The set Q consists of the following terms: 18.29/5.51 18.29/5.51 3(4(5(x0))) 18.29/5.51 18.29/5.51 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (5) DependencyPairsProof (EQUIVALENT) 18.29/5.51 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 18.29/5.51 ---------------------------------------- 18.29/5.51 18.29/5.51 (6) 18.29/5.51 Obligation: 18.29/5.51 Q DP problem: 18.29/5.51 The TRS P consists of the following rules: 18.29/5.51 18.29/5.51 3^1(4(5(x1))) -> 3^1(5(x1)) 18.29/5.51 18.29/5.51 The TRS R consists of the following rules: 18.29/5.51 18.29/5.51 3(4(5(x1))) -> 4(3(5(x1))) popout

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all output return to SRS_Standard 2019-03-29 03.29