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TRS Stand 20472 pair #381712319
details
property
value
status
complete
benchmark
5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n028.star.cs.uiowa.edu
space
Secret_06_TRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
2.03332901001 seconds
cpu usage
4.75578263
max memory
3.28531968E8
stage attributes
key
value
output-size
4815
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 62 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) MRRProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 37 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(y, 0), 0) -> y c(c(y)) -> y c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(a(x_1, x_2)) = x_1 + x_2 POL(c(x_1)) = 2 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: c(c(y)) -> y ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(y, 0), 0) -> y c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y) Q is empty. ---------------------------------------- (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: C(a(c(c(y)), x)) -> A(c(c(c(a(x, 0)))), y) C(a(c(c(y)), x)) -> C(c(c(a(x, 0)))) C(a(c(c(y)), x)) -> C(c(a(x, 0))) C(a(c(c(y)), x)) -> C(a(x, 0)) C(a(c(c(y)), x)) -> A(x, 0) The TRS R consists of the following rules: a(a(y, 0), 0) -> y c(a(c(c(y)), x)) -> a(c(c(c(a(x, 0)))), y) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6)
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