details

property	value
status	complete
benchmark	Ex1_GM99_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.86122 seconds
cpu usage	4.08783
user time	3.89492
system time	0.192908
max virtual memory	1.841096E7
max residence set size	299612.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) NonTerminationLoopProof [COMPLETE, 0 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(n__a, n__b, X) -> f(X, X, X) c -> a c -> b a -> n__a b -> n__b activate(n__a) -> a activate(n__b) -> b activate(X) -> 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(n__a, n__b, X) -> F(X, X, X) C -> A C -> B ACTIVATE(n__a) -> A ACTIVATE(n__b) -> B The TRS R consists of the following rules: f(n__a, n__b, X) -> f(X, X, X) c -> a c -> b a -> n__a b -> n__b activate(n__a) -> a activate(n__b) -> b activate(X) -> 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 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(n__a, n__b, X) -> F(X, X, X) The TRS R consists of the following rules: f(n__a, n__b, X) -> f(X, X, X) c -> a c -> b a -> n__a b -> n__b activate(n__a) -> a activate(n__b) -> b activate(X) -> X 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(c, c, X) evaluates to t =F(X, X, X) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] popout

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all output return to TRS Standard