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TRS Inner 89993 pair #381733566
details
property
value
status
complete
benchmark
#4.16.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n030.star.cs.uiowa.edu
space
AG01_innermost
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.61853504181 seconds
cpu usage
3.452255599
max memory
2.28220928E8
stage attributes
key
value
output-size
3449
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(0), g(x)) -> f(x, g(x)) g(s(x)) -> g(x) The set Q consists of the following terms: f(s(0), g(x0)) g(s(x0)) ---------------------------------------- (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(s(0), g(x)) -> F(x, g(x)) G(s(x)) -> G(x) The TRS R consists of the following rules: f(s(0), g(x)) -> f(x, g(x)) g(s(x)) -> g(x) The set Q consists of the following terms: f(s(0), g(x0)) g(s(x0)) 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: G(s(x)) -> G(x) The TRS R consists of the following rules: f(s(0), g(x)) -> f(x, g(x)) g(s(x)) -> g(x) The set Q consists of the following terms: f(s(0), g(x0)) g(s(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem:
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