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SRS Standard pair #487086730
details
property
value
status
complete
benchmark
4002.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
ICFP_2010
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
11.2895 seconds
cpu usage
41.3053
user time
39.5325
system time
1.77283
max virtual memory
4.0137192E7
max residence set size
4153488.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 29 ms] (4) QTRS (5) QTRSRRRProof [EQUIVALENT, 2643 ms] (6) QTRS (7) QTRSRRRProof [EQUIVALENT, 20 ms] (8) QTRS (9) QTRSRRRProof [EQUIVALENT, 22 ms] (10) QTRS (11) QTRSRRRProof [EQUIVALENT, 11 ms] (12) QTRS (13) QTRSRRRProof [EQUIVALENT, 2 ms] (14) QTRS (15) RisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 2(5(3(0(x1)))) -> 1(0(0(1(3(0(4(5(1(2(x1)))))))))) 1(3(5(4(3(x1))))) -> 2(1(4(1(4(0(3(0(1(1(x1)))))))))) 5(1(3(5(0(x1))))) -> 5(1(4(3(0(4(4(5(2(1(x1)))))))))) 5(4(4(2(5(x1))))) -> 4(3(1(1(1(1(5(3(3(5(x1)))))))))) 2(2(5(0(5(4(x1)))))) -> 2(1(4(1(3(3(2(2(5(4(x1)))))))))) 3(0(5(5(4(3(x1)))))) -> 3(3(0(3(2(3(5(5(1(0(x1)))))))))) 3(5(4(3(5(2(x1)))))) -> 2(0(5(2(0(5(2(2(3(2(x1)))))))))) 4(4(2(5(5(0(x1)))))) -> 4(4(0(0(3(3(3(2(2(3(x1)))))))))) 4(5(3(5(5(0(x1)))))) -> 4(2(2(3(0(2(4(1(1(5(x1)))))))))) 5(4(5(1(1(2(x1)))))) -> 5(4(0(3(3(3(3(2(5(5(x1)))))))))) 5(5(5(5(5(3(x1)))))) -> 5(5(0(1(4(0(0(5(0(1(x1)))))))))) 3(5(0(0(5(4(3(x1))))))) -> 0(1(2(1(1(5(5(2(1(0(x1)))))))))) 3(5(4(2(5(2(3(x1))))))) -> 4(0(4(0(0(2(2(3(4(4(x1)))))))))) 3(5(4(5(1(4(0(x1))))))) -> 1(1(1(0(0(3(3(1(2(5(x1)))))))))) Q is empty. ---------------------------------------- (1) FlatCCProof (EQUIVALENT) We used flat context closure [ROOTLAB] As Q is empty the flat context closure was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 5(1(3(5(0(x1))))) -> 5(1(4(3(0(4(4(5(2(1(x1)))))))))) 2(2(5(0(5(4(x1)))))) -> 2(1(4(1(3(3(2(2(5(4(x1)))))))))) 3(0(5(5(4(3(x1)))))) -> 3(3(0(3(2(3(5(5(1(0(x1)))))))))) 4(4(2(5(5(0(x1)))))) -> 4(4(0(0(3(3(3(2(2(3(x1)))))))))) 4(5(3(5(5(0(x1)))))) -> 4(2(2(3(0(2(4(1(1(5(x1)))))))))) 5(4(5(1(1(2(x1)))))) -> 5(4(0(3(3(3(3(2(5(5(x1)))))))))) 5(5(5(5(5(3(x1)))))) -> 5(5(0(1(4(0(0(5(0(1(x1)))))))))) 2(2(5(3(0(x1))))) -> 2(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 5(2(5(3(0(x1))))) -> 5(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 3(2(5(3(0(x1))))) -> 3(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 0(2(5(3(0(x1))))) -> 0(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 1(2(5(3(0(x1))))) -> 1(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 4(2(5(3(0(x1))))) -> 4(1(0(0(1(3(0(4(5(1(2(x1))))))))))) 2(1(3(5(4(3(x1)))))) -> 2(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 5(1(3(5(4(3(x1)))))) -> 5(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 3(1(3(5(4(3(x1)))))) -> 3(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 0(1(3(5(4(3(x1)))))) -> 0(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 1(1(3(5(4(3(x1)))))) -> 1(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 4(1(3(5(4(3(x1)))))) -> 4(2(1(4(1(4(0(3(0(1(1(x1))))))))))) 2(5(4(4(2(5(x1)))))) -> 2(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 5(5(4(4(2(5(x1)))))) -> 5(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 3(5(4(4(2(5(x1)))))) -> 3(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 0(5(4(4(2(5(x1)))))) -> 0(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 1(5(4(4(2(5(x1)))))) -> 1(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 4(5(4(4(2(5(x1)))))) -> 4(4(3(1(1(1(1(5(3(3(5(x1))))))))))) 2(3(5(4(3(5(2(x1))))))) -> 2(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 5(3(5(4(3(5(2(x1))))))) -> 5(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 3(3(5(4(3(5(2(x1))))))) -> 3(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 0(3(5(4(3(5(2(x1))))))) -> 0(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 1(3(5(4(3(5(2(x1))))))) -> 1(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 4(3(5(4(3(5(2(x1))))))) -> 4(2(0(5(2(0(5(2(2(3(2(x1))))))))))) 2(3(5(0(0(5(4(3(x1)))))))) -> 2(0(1(2(1(1(5(5(2(1(0(x1))))))))))) 5(3(5(0(0(5(4(3(x1)))))))) -> 5(0(1(2(1(1(5(5(2(1(0(x1))))))))))) 3(3(5(0(0(5(4(3(x1)))))))) -> 3(0(1(2(1(1(5(5(2(1(0(x1))))))))))) 0(3(5(0(0(5(4(3(x1)))))))) -> 0(0(1(2(1(1(5(5(2(1(0(x1))))))))))) 1(3(5(0(0(5(4(3(x1)))))))) -> 1(0(1(2(1(1(5(5(2(1(0(x1))))))))))) 4(3(5(0(0(5(4(3(x1)))))))) -> 4(0(1(2(1(1(5(5(2(1(0(x1)))))))))))
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