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TRS Stand 20472 pair #381715078
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
n104.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.96208405495 seconds
cpu usage
4.307927837
max memory
3.08293632E8
stage attributes
key
value
output-size
4837
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 58 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesReductionPairsProof [EQUIVALENT, 7 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
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