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TRS Standard pair #487072888
details
property
value
status
complete
benchmark
ex2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n186.star.cs.uiowa.edu
space
Strategy_removed_mixed_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.10889 seconds
cpu usage
5.0587
user time
4.84617
system time
0.212523
max virtual memory
3.684186E7
max residence set size
374292.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) NonTerminationLoopProof [COMPLETE, 0 ms] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(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: INF(x) -> CONS(x, inf(s(x))) INF(x) -> INF(s(x)) The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: INF(x) -> INF(s(x)) The TRS R consists of the following rules: cons(x, cons(y, z)) -> big inf(x) -> cons(x, inf(s(x))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: INF(x) -> INF(s(x)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) TransformationProof (EQUIVALENT) By instantiating [LPAR04] the rule INF(x) -> INF(s(x)) we obtained the following new rules [LPAR04]: (INF(s(z0)) -> INF(s(s(z0))),INF(s(z0)) -> INF(s(s(z0)))) ----------------------------------------
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