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TRS Stand 20472 pair #381712678
details
property
value
status
complete
benchmark
matchbox2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
Secret_05_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.89583706856 seconds
cpu usage
4.167672079
max memory
2.51334656E8
stage attributes
key
value
output-size
3403
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 17 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(f(x, a), y) -> f(f(a, y), f(a, 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: F(f(x, a), y) -> F(f(a, y), f(a, x)) F(f(x, a), y) -> F(a, y) F(f(x, a), y) -> F(a, x) The TRS R consists of the following rules: f(f(x, a), y) -> f(f(a, y), f(a, 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 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(f(x, a), y) -> F(f(a, y), f(a, x)) The TRS R consists of the following rules: f(f(x, a), y) -> f(f(a, y), f(a, 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: F(f(x, a), y) -> F(f(a, y), f(a, x)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted.
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