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TRS Standard pair #516964070
details
property
value
status
complete
benchmark
02.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n173.star.cs.uiowa.edu
space
Various_04
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.11120510101 seconds
cpu usage
5.294293198
max memory
4.76672E8
stage attributes
key
value
output-size
1494
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(x, x, x) f(x, y, z) -> 2 0 -> 2 1 -> 2 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(0, 1, x) -> F(x, x, x) The TRS R consists of the following rules: f(0, 1, x) -> f(x, x, x) f(x, y, z) -> 2 0 -> 2 1 -> 2 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 0 SCCs with 1 less node. ---------------------------------------- (4) TRUE
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