details

property	value
status	complete
benchmark	Ex24_GM04.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n112.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.04472589493 seconds
cpu usage	4.203357106
max memory	2.46407168E8

stage attributes

key	value
output-size	2984
starexec-result	NO

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) TransformationProof [EQUIVALENT, 0 ms] (4) QDP (5) NonTerminationLoopProof [COMPLETE, 21 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) g(b) -> c b -> c 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, g(X), Y) -> F(Y, Y, Y) The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) g(b) -> c b -> c Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) TransformationProof (EQUIVALENT) By forward instantiating [JAR06] the rule F(X, g(X), Y) -> F(Y, Y, Y) we obtained the following new rules [LPAR04]: (F(x0, g(x0), g(y_1)) -> F(g(y_1), g(y_1), g(y_1)),F(x0, g(x0), g(y_1)) -> F(g(y_1), g(y_1), g(y_1))) ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(x0, g(x0), g(y_1)) -> F(g(y_1), g(y_1), g(y_1)) The TRS R consists of the following rules: f(X, g(X), Y) -> f(Y, Y, Y) g(b) -> c b -> c Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = F(g(b), g(b), g(y_1)) evaluates to t =F(g(y_1), g(y_1), g(y_1)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [y_1 / b] -------------------------------------------------------------------------------- Rewriting sequence F(g(b), g(b), g(b)) -> F(g(b), g(c), g(b)) with rule b -> c at position [1,0] and matcher [ ] F(g(b), g(c), g(b)) -> F(c, g(c), g(b)) popout

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