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	n071.star.cs.uiowa.edu
space	Waldmann_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.50526 seconds
cpu usage	6.69472
user time	6.41743
system time	0.27729
max virtual memory	1.8484924E7
max residence set size	391244.0

stage attributes

key	value
starexec-result	NO

output

NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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, 15 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: 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. ---------------------------------------- popout

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actions

all output return to TRS Standard