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TRS Stand 20472 pair #381715677
details
property
value
status
complete
benchmark
Ex9_BLR02_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.6440179348 seconds
cpu usage
3.624992013
max memory
1.90943232E8
stage attributes
key
value
output-size
1835
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 30 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: filter(cons(X), 0, M) -> cons(0) filter(cons(X), s(N), M) -> cons(X) sieve(cons(0)) -> cons(0) sieve(cons(s(N))) -> cons(s(N)) nats(N) -> cons(N) zprimes -> sieve(nats(s(s(0)))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:sieve_1 > zprimes > s_1 > cons_1 > nats_1 > 0 > filter_3 and weight map: 0=3 zprimes=8 cons_1=1 s_1=1 sieve_1=0 nats_1=2 filter_3=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: filter(cons(X), 0, M) -> cons(0) filter(cons(X), s(N), M) -> cons(X) sieve(cons(0)) -> cons(0) sieve(cons(s(N))) -> cons(s(N)) nats(N) -> cons(N) zprimes -> sieve(nats(s(s(0)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES
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