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TRS Standard pair #487071463
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
n187.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.56583 seconds
cpu usage
6.97793
user time
6.69241
system time
0.285521
max virtual memory
1.848154E7
max residence set size
441796.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 40 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, 140 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) 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))
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