details

property	value
status	complete
benchmark	247992.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.9208 seconds
cpu usage	8.60729
user time	8.23629
system time	0.371007
max virtual memory	1.907558E7
max residence set size	1128468.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 146 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 2 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 10 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(x1) -> 1(x1) 0(0(x1)) -> 0(x1) 3(4(5(x1))) -> 4(3(5(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(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(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(1(0(1(1(1(1(0(0(0(0(0(0(1(1(1(1(0(0(0(0(1(0(0(0(0(1(1(1(0(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(1(0(0(0(0(1(1(0(0(1(0(0(0(0(0(0(1(0(0(0(1(1(1(0(1(0(0(0(1(1(0(0(1(0(0(0(0(0(1(0(0(1(1(1(0(1(0(1(1(1(0(1(0(1(1(1(0(1(0(0(1(0(1(1(1(1(0(1(0(1(0(1(1(0(1(0(0(1(0(0(0(1(0(1(1(1(1(0(1(0(1(0(1(1(0(0(0(0(0(1(0(0(0(1(0(1(1(1(1(0(1(1(0(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(1(0(0(1(1(0(0(0(1(0(0(0(1(1(0(0(1(0(0(1(1(0(0(1(0(0(0(1(0(1(0(0(0(0(0(0(0(0(1(1(0(0(1(0(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(1(0(1(1(0(0(1(0(1(1(0(1(1(0(1(1(1(0(1(0(0(0(1(1(1(0(1(0(1(0(1(0(1(1(1(1(1(0(1(0(0(1(1(1(1(0(1(0(1(1(1(1(0(1(1(1(0(1(1(1(0(0(1(0(1(0(1(0(0(1(1(1(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 0(1(0(1(0(1(0(0(0(0(1(0(0(1(1(0(1(1(0(0(1(0(0(0(1(0(1(1(1(1(1(1(1(1(0(0(0(1(1(1(0(0(1(0(0(1(0(0(0(1(1(0(0(0(0(1(0(0(1(1(0(1(0(1(1(1(1(1(1(1(1(1(0(1(1(0(1(0(0(0(1(0(0(1(0(1(1(1(1(1(1(1(0(0(0(0(1(1(0(1(1(0(0(1(0(0(1(0(1(0(1(1(0(0(0(1(1(0(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(0(1(0(0(1(1(1(0(1(1(0(0(1(0(0(1(1(1(0(1(0(0(1(0(1(1(0(1(1(0(1(1(0(0(1(0(1(0(0(0(0(1(1(0(0(0(1(1(0(0(1(0(1(1(1(1(1(1(1(1(0(0(0(0(0(1(0(1(0(1(1(1(0(1(0(1(1(1(1(1(1(1(0(1(1(0(1(0(0(0(1(1(1(0(1(1(0(0(0(0(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(0(0(1(0(1(0(1(0(0(0(1(0(1(1(1(0(0(1(1(0(0(1(1(0(0(0(1(1(1(1(1(1(0(0(1(0(0(0(0(1(0(1(1(0(1(1(1(0(0(0(1(0(1(0(1(0(1(0(0(0(1(0(0(0(1(0(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(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(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))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 12 + x_1 POL(1(x_1)) = 11 + x_1 POL(2(x_1)) = 24 + x_1 POL(3(x_1)) = x_1 POL(4(x_1)) = x_1 POL(5(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 0(x1) -> 1(x1) 0(0(x1)) -> 0(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(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(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(1(0(1(1(1(1(0(0(0(0(0(0(1(1(1(1(0(0(0(0(1(0(0(0(0(1(1(1(0(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(1(0(0(0(0(1(1(0(0(1(0(0(0(0(0(0(1(0(0(0(1(1(1(0(1(0(0(0(1(1(0(0(1(0(0(0(0(0(1(0(0(1(1(1(0(1(0(1(1(1(0(1(0(1(1(1(0(1(0(0(1(0(1(1(1(1(0(1(0(1(0(1(1(0(1(0(0(1(0(0(0(1(0(1(1(1(1(0(1(0(1(0(1(1(0(0(0(0(0(1(0(0(0(1(0(1(1(1(1(0(1(1(0(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(1(0(0(1(1(0(0(0(1(0(0(0(1(1(0(0(1(0(0(1(1(0(0(1(0(0(0(1(0(1(0(0(0(0(0(0(0(0(1(1(0(0(1(0(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(1(0(1(1(0(0(1(0(1(1(0(1(1(0(1(1(1(0(1(0(0(0(1(1(1(0(1(0(1(0(1(0(1(1(1(1(1(0(1(0(0(1(1(1(1(0(1(0(1(1(1(1(0(1(1(1(0(1(1(1(0(0(1(0(1(0(1(0(0(1(1(1(1(0(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 0(1(0(1(0(1(0(0(0(0(1(0(0(1(1(0(1(1(0(0(1(0(0(0(1(0(1(1(1(1(1(1(1(1(0(0(0(1(1(1(0(0(1(0(0(1(0(0(0(1(1(0(0(0(0(1(0(0(1(1(0(1(0(1(1(1(1(1(1(1(1(1(0(1(1(0(1(0(0(0(1(0(0(1(0(1(1(1(1(1(1(1(0(0(0(0(1(1(0(1(1(0(0(1(0(0(1(0(1(0(1(1(0(0(0(1(1(0(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(0(1(0(0(1(1(1(0(1(1(0(0(1(0(0(1(1(1(0(1(0(0(1(0(1(1(0(1(1(0(1(1(0(0(1(0(1(0(0(0(0(1(1(0(0(0(1(1(0(0(1(0(1(1(1(1(1(1(1(1(0(0(0(0(0(1(0(1(0(1(1(1(0(1(0(1(1(1(1(1(1(1(0(1(1(0(1(0(0(0(1(1(1(0(1(1(0(0(0(0(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(0(0(1(0(1(0(1(0(0(0(1(0(1(1(1(0(0(1(1(0(0(1(1(0(0(0(1(1(1(1(1(1(0(0(1(0(0(0(0(1(0(1(1(0(1(1(1(0(0(0(1(0(1(0(1(0(1(0(0(0(1(0(0(0(1(0(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(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(2(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))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) The set Q consists of the following terms: 3(4(5(x0))) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 3^1(4(5(x1))) -> 3^1(5(x1)) The TRS R consists of the following rules: 3(4(5(x1))) -> 4(3(5(x1))) popout

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