details

property	value
status	complete
benchmark	Ex24_Luc06_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)	1.93896 seconds
cpu usage	4.21903
user time	3.99854
system time	0.220493
max virtual memory	1.8411988E7
max residence set size	315488.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 58 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) NonTerminationLoopProof [COMPLETE, 0 ms] (10) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(n__b, X, n__c) -> f(X, c, X) c -> b b -> n__b c -> n__c activate(n__b) -> b activate(n__c) -> c activate(X) -> X Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(activate(x_1)) = 2 + 2*x_1 POL(b) = 0 POL(c) = 0 POL(f(x_1, x_2, x_3)) = x_1 + 2*x_2 + x_3 POL(n__b) = 0 POL(n__c) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: activate(n__b) -> b activate(n__c) -> c activate(X) -> X ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(n__b, X, n__c) -> f(X, c, X) c -> b b -> n__b c -> n__c Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(n__b, X, n__c) -> F(X, c, X) F(n__b, X, n__c) -> C C -> B The TRS R consists of the following rules: f(n__b, X, n__c) -> f(X, c, X) c -> b b -> n__b c -> n__c Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. popout

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job log

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actions

all output return to TRS Standard