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SRS Stand 10685 pair #381717665
details
property
value
status
complete
benchmark
jambox5.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n027.star.cs.uiowa.edu
space
Secret_05_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
15.9181680679 seconds
cpu usage
59.671109672
max memory
5.086490624E9
stage attributes
key
value
output-size
10284
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 7 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) MRRProof [EQUIVALENT, 16 ms] (11) QDP (12) DependencyGraphProof [EQUIVALENT, 0 ms] (13) TRUE (14) QDP (15) QDPOrderProof [EQUIVALENT, 115 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 776 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) QDP (21) UsableRulesProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(x1)) -> a(a(d(x1))) a(c(x1)) -> b(b(x1)) d(a(b(x1))) -> b(d(d(c(x1)))) d(x1) -> a(x1) b(a(c(a(x1)))) -> x1 Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> d(a(a(x1))) c(a(x1)) -> b(b(x1)) b(a(d(x1))) -> c(d(d(b(x1)))) d(x1) -> a(x1) a(c(a(b(x1)))) -> 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(b(x1)) -> D(a(a(x1))) A(b(x1)) -> A(a(x1)) A(b(x1)) -> A(x1) C(a(x1)) -> B(b(x1)) C(a(x1)) -> B(x1) B(a(d(x1))) -> C(d(d(b(x1)))) B(a(d(x1))) -> D(d(b(x1))) B(a(d(x1))) -> D(b(x1)) B(a(d(x1))) -> B(x1) D(x1) -> A(x1) The TRS R consists of the following rules: a(b(x1)) -> d(a(a(x1))) c(a(x1)) -> b(b(x1))
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