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SRS Stand 10685 pair #381718971
details
property
value
status
complete
benchmark
z083.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
4.27441382408 seconds
cpu usage
12.857767468
max memory
1.449951232E9
stage attributes
key
value
output-size
6388
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 41 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 1 ms] (8) QDP (9) MRRProof [EQUIVALENT, 38 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 54 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(d(b(x1))) -> c(d(b(x1))) b(a(c(x1))) -> b(c(x1)) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(x1)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 POL(d(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: b(a(c(x1))) -> b(c(x1)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(d(b(x1))) -> c(d(b(x1))) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(x1)) 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: A(d(x1)) -> D(c(x1)) B(b(b(x1))) -> A(b(c(x1))) B(b(b(x1))) -> B(c(x1)) D(c(x1)) -> B(d(x1)) D(c(x1)) -> D(x1) D(c(x1)) -> D(b(d(x1))) D(a(c(x1))) -> B(b(x1)) D(a(c(x1))) -> B(x1)
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