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TRS Stand 20472 pair #381716382
details
property
value
status
complete
benchmark
Ex9_Luc04_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n105.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.57266902924 seconds
cpu usage
6.90729375
max memory
3.756032E8
stage attributes
key
value
output-size
9905
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 25 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) NonTerminationLoopProof [COMPLETE, 84 ms] (16) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(a, b, X) -> a__f(mark(X), X, mark(X)) a__c -> a a__c -> b mark(f(X1, X2, X3)) -> a__f(mark(X1), X2, mark(X3)) mark(c) -> a__c mark(a) -> a mark(b) -> b a__f(X1, X2, X3) -> f(X1, X2, X3) a__c -> 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: A__F(a, b, X) -> A__F(mark(X), X, mark(X)) A__F(a, b, X) -> MARK(X) MARK(f(X1, X2, X3)) -> A__F(mark(X1), X2, mark(X3)) MARK(f(X1, X2, X3)) -> MARK(X1) MARK(f(X1, X2, X3)) -> MARK(X3) MARK(c) -> A__C The TRS R consists of the following rules: a__f(a, b, X) -> a__f(mark(X), X, mark(X)) a__c -> a a__c -> b mark(f(X1, X2, X3)) -> a__f(mark(X1), X2, mark(X3)) mark(c) -> a__c mark(a) -> a mark(b) -> b a__f(X1, X2, X3) -> f(X1, X2, X3) a__c -> c 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: A__F(a, b, X) -> MARK(X) MARK(f(X1, X2, X3)) -> A__F(mark(X1), X2, mark(X3)) A__F(a, b, X) -> A__F(mark(X), X, mark(X)) MARK(f(X1, X2, X3)) -> MARK(X1) MARK(f(X1, X2, X3)) -> MARK(X3)
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