details

property	value
status	complete
benchmark	211471.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	ICFP_2010

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.0801 seconds
cpu usage	12.9465
user time	11.6822
system time	1.26426
max virtual memory	1.9017344E7
max residence set size	2811424.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 62 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 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(x1))))))))))))) -> 0(0(0(1(0(1(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(0(0(1(1(0(0(0(1(0(1(0(1(0(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(0(0(0(1(1(1(1(0(1(0(1(0(0(0(0(1(1(0(0(1(1(1(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 1(1(0(0(1(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(1(1(0(0(1(0(1(0(0(0(0(0(0(1(0(0(0(1(0(1(1(0(0(1(1(0(1(1(0(1(1(0(1(0(0(1(1(1(0(1(0(0(1(0(0(0(0(0(0(1(1(1(0(1(1(0(0(1(1(0(1(0(0(1(1(0(0(1(0(1(1(0(1(0(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(0(1(1(0(1(1(0(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 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)) = 2 + x_1 POL(1(x_1)) = 1 + x_1 POL(2(x_1)) = 17 + 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(x1))))))))))))) -> 0(0(0(1(0(1(1(1(0(1(1(1(1(0(0(0(0(1(0(0(1(1(1(1(0(0(0(0(0(0(0(0(1(0(0(0(1(1(1(1(1(1(1(1(1(0(0(0(0(1(1(1(1(1(1(1(0(0(0(1(0(1(1(1(0(0(0(1(1(1(1(1(1(1(1(0(1(0(0(1(0(0(1(1(0(0(0(1(0(1(0(1(0(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(0(0(0(0(1(0(0(0(0(0(0(0(0(0(1(1(1(1(0(1(0(1(0(0(0(0(1(1(0(0(1(1(1(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) 1(1(0(0(1(1(0(1(0(0(1(0(1(1(1(1(1(0(0(1(0(1(1(1(0(1(0(0(0(0(1(0(1(1(1(1(1(0(0(1(0(0(1(1(0(0(0(0(1(1(0(0(1(0(1(0(0(0(0(0(0(1(0(0(0(1(0(1(1(0(0(1(1(0(1(1(0(1(1(0(1(0(0(1(1(1(0(1(0(0(1(0(0(0(0(0(0(1(1(1(0(1(1(0(0(1(1(0(1(0(0(1(1(0(0(1(0(1(1(0(1(0(1(1(1(0(1(0(1(1(0(0(0(1(0(1(1(0(1(1(0(1(1(0(0(0(1(1(x1)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) -> 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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