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TRS Stand 20472 pair #381710471
details
property
value
status
complete
benchmark
forward_instantiation2.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
AProVE_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.56397414207 seconds
cpu usage
4.19795829
max memory
2.52760064E8
stage attributes
key
value
output-size
3942
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y, z) -> g(x, y, z) g(0, 1, x) -> f(x, x, x) a -> b a -> c Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is f(x, y, z) -> g(x, y, z) g(0, 1, x) -> f(x, x, x) a -> b a -> c The signature Sigma is {f_3, g_3, a, b, c} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, y, z) -> g(x, y, z) g(0, 1, x) -> f(x, x, x) a -> b a -> c The set Q consists of the following terms: f(x0, x1, x2) g(0, 1, x0) a ---------------------------------------- (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: F(x, y, z) -> G(x, y, z) G(0, 1, x) -> F(x, x, x) The TRS R consists of the following rules: f(x, y, z) -> g(x, y, z) g(0, 1, x) -> f(x, x, x) a -> b a -> c The set Q consists of the following terms: f(x0, x1, x2) g(0, 1, x0) a
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