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TRS Stand 20472 pair #381709642
details
property
value
status
complete
benchmark
ex3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n020.star.cs.uiowa.edu
space
Strategy_removed_mixed_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.9110558033 seconds
cpu usage
8.454432545
max memory
1.031917568E9
stage attributes
key
value
output-size
3538
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 16 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 14 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [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(g(a)) -> f(s(g(b))) f(f(x)) -> b g(x) -> f(g(x)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a) = 1 POL(b) = 0 POL(f(x_1)) = x_1 POL(g(x_1)) = x_1 POL(s(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(g(a)) -> f(s(g(b))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(x)) -> b g(x) -> f(g(x)) 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: G(x) -> F(g(x)) G(x) -> G(x) The TRS R consists of the following rules: f(f(x)) -> b g(x) -> f(g(x)) 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 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules:
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