details

property	value
status	complete
benchmark	aabacaaa-aaaaabacaabac.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n050.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	175.438272953 seconds
cpu usage	693.422169529
max memory	1.5463829504E10

stage attributes

key	value
output-size	5915
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 13 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 8703 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 6199 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(a(c(a(a(a(x1)))))))) -> a(a(a(a(a(b(a(c(a(a(b(a(c(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(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(a(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(a(a(c(a(b(a(a(x1)))))))) -> A(b(a(a(c(a(b(a(a(a(a(a(x1)))))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(c(a(b(a(a(a(a(a(x1)))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(c(a(b(a(a(a(a(a(x1))))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(b(a(a(a(a(a(x1))))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(a(x1))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(x1)))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(x1))) The TRS R consists of the following rules: a(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(a(x1))))))))))))) 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 4 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(x1)))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(a(a(x1))))) A(a(a(c(a(b(a(a(x1)))))))) -> A(a(a(x1))) The TRS R consists of the following rules: a(a(a(c(a(b(a(a(x1)))))))) -> c(a(b(a(a(c(a(b(a(a(a(a(a(x1))))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Stand 10685