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TRS Stand 20472 pair #381710310
details
property
value
status
complete
benchmark
7.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n058.star.cs.uiowa.edu
space
Secret_07_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
125.67162323 seconds
cpu usage
383.823610766
max memory
1.0063089664E10
stage attributes
key
value
output-size
10223
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) NonTerminationLoopProof [COMPLETE, 77.0 s] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(a, f(a, f(b, f(x, y)))) -> f(b, f(c, f(b, f(a, f(a, f(a, f(x, y))))))) f(a, f(c, f(x, y))) -> f(b, f(x, y)) 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: F(a, f(a, f(b, f(x, y)))) -> F(b, f(c, f(b, f(a, f(a, f(a, f(x, y))))))) F(a, f(a, f(b, f(x, y)))) -> F(c, f(b, f(a, f(a, f(a, f(x, y)))))) F(a, f(a, f(b, f(x, y)))) -> F(b, f(a, f(a, f(a, f(x, y))))) F(a, f(a, f(b, f(x, y)))) -> F(a, f(a, f(a, f(x, y)))) F(a, f(a, f(b, f(x, y)))) -> F(a, f(a, f(x, y))) F(a, f(a, f(b, f(x, y)))) -> F(a, f(x, y)) F(a, f(c, f(x, y))) -> F(b, f(x, y)) The TRS R consists of the following rules: f(a, f(a, f(b, f(x, y)))) -> f(b, f(c, f(b, f(a, f(a, f(a, f(x, y))))))) f(a, f(c, f(x, y))) -> f(b, f(x, y)) 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 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(a, f(a, f(b, f(x, y)))) -> F(a, f(a, f(x, y))) F(a, f(a, f(b, f(x, y)))) -> F(a, f(a, f(a, f(x, y)))) F(a, f(a, f(b, f(x, y)))) -> F(a, f(x, y)) The TRS R consists of the following rules: f(a, f(a, f(b, f(x, y)))) -> f(b, f(c, f(b, f(a, f(a, f(a, f(x, y))))))) f(a, f(c, f(x, y))) -> f(b, f(x, y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule F(a, f(a, f(b, f(x, y)))) -> F(a, f(a, f(a, f(x, y)))) at position [1] we obtained the following new rules [LPAR04]: (F(a, f(a, f(b, f(b, f(x0, x1))))) -> F(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1)))))))),F(a, f(a, f(b, f(b, f(x0, x1))))) -> F(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1))))))))) (F(a, f(a, f(b, f(a, f(b, f(x0, x1)))))) -> F(a, f(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1))))))))),F(a, f(a, f(b, f(a, f(b, f(x0, x1)))))) -> F(a, f(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1)))))))))) (F(a, f(a, f(b, f(c, f(x0, x1))))) -> F(a, f(a, f(b, f(x0, x1)))),F(a, f(a, f(b, f(c, f(x0, x1))))) -> F(a, f(a, f(b, f(x0, x1))))) (F(a, f(a, f(b, f(a, f(a, f(b, f(x0, x1))))))) -> F(a, f(a, f(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1)))))))))),F(a, f(a, f(b, f(a, f(a, f(b, f(x0, x1))))))) -> F(a, f(a, f(a, f(b, f(c, f(b, f(a, f(a, f(a, f(x0, x1))))))))))) (F(a, f(a, f(b, f(a, f(c, f(x0, x1)))))) -> F(a, f(a, f(a, f(b, f(x0, x1))))),F(a, f(a, f(b, f(a, f(c, f(x0, x1)))))) -> F(a, f(a, f(a, f(b, f(x0, x1))))))
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