details

property	value
status	complete
benchmark	4.49.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n099.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.90778398514 seconds
cpu usage	4.540622647
max memory	3.10272E8

stage attributes

key	value
output-size	2079
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 0 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y, f(z, u, v)) -> f(f(x, y, z), u, f(x, y, v)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, y, f(z, u, v)) -> F(f(x, y, z), u, f(x, y, v)) F(x, y, f(z, u, v)) -> F(x, y, z) F(x, y, f(z, u, v)) -> F(x, y, v) The TRS R consists of the following rules: f(x, y, f(z, u, v)) -> f(f(x, y, z), u, f(x, y, v)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = F(x, y, f(z, u, v)) evaluates to t =F(f(x, y, z), u, f(x, y, v)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x / f(x, y, z), y / u, z / x, u / y] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from F(x, y, f(z, u, v)) to F(f(x, y, z), u, f(x, y, v)). ---------------------------------------- (4) NO popout

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all output return to TRS Stand 20472