details

property	value
status	complete
benchmark	jwno9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n035.star.cs.uiowa.edu
space	Waldmann_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.61159777641 seconds
cpu usage	6.545400431
max memory	4.20864E8

stage attributes

key	value
output-size	4777
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) NonTerminationLoopProof [COMPLETE, 0 ms] (8) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, f(a, y)) -> f(a, f(f(f(a, a), y), x)) 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, f(a, y)) -> F(a, f(f(f(a, a), y), x)) F(x, f(a, y)) -> F(f(f(a, a), y), x) F(x, f(a, y)) -> F(f(a, a), y) F(x, f(a, y)) -> F(a, a) The TRS R consists of the following rules: f(x, f(a, y)) -> f(a, f(f(f(a, a), y), x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, f(a, y)) -> F(f(f(a, a), y), x) F(x, f(a, y)) -> F(a, f(f(f(a, a), y), x)) F(x, f(a, y)) -> F(f(a, a), y) The TRS R consists of the following rules: f(x, f(a, y)) -> f(a, f(f(f(a, a), y), x)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(x, f(a, y)) -> F(a, f(f(f(a, a), y), x)) at position [1] we obtained the following new rules [LPAR04]: (F(f(a, x1), f(a, y1)) -> F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1)))),F(f(a, x1), f(a, y1)) -> F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1))))) (F(y0, f(a, f(a, x1))) -> F(a, f(f(a, f(f(f(a, a), x1), f(a, a))), y0)),F(y0, f(a, f(a, x1))) -> F(a, f(f(a, f(f(f(a, a), x1), f(a, a))), y0))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, f(a, y)) -> F(f(f(a, a), y), x) F(x, f(a, y)) -> F(f(a, a), y) F(f(a, x1), f(a, y1)) -> F(a, f(a, f(f(f(a, a), x1), f(f(a, a), y1)))) F(y0, f(a, f(a, x1))) -> F(a, f(f(a, f(f(f(a, a), x1), f(a, a))), y0)) The TRS R consists of the following rules: popout

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

job log

popout

actions

all output return to TRS Stand 20472