details

property	value
status	complete
benchmark	un14.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	38.1353 seconds
cpu usage	146.988
user time	142.839
system time	4.14897
max virtual memory	8.3171912E7
max residence set size	8500372.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 23 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 ms] (6) AND (7) QDP (8) QDPOrderProof [EQUIVALENT, 5558 ms] (9) QDP (10) PisEmptyProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 5917 ms] (14) QDP (15) PisEmptyProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(b(x1))) -> b(a(a(a(b(x1))))) b(a(a(a(b(a(a(b(x1)))))))) -> b(a(a(b(a(a(b(a(a(a(b(b(x1)))))))))))) b(a(a(a(b(a(a(a(b(x1))))))))) -> b(x1) b(a(a(a(b(b(b(x1))))))) -> b(b(b(a(a(a(b(x1))))))) b(a(a(b(b(x1))))) -> b(x1) b(b(a(a(b(x1))))) -> b(x1) b(a(a(a(b(a(b(x1))))))) -> b(x1) b(a(b(a(a(a(b(x1))))))) -> b(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: b(a(b(x1))) -> b(a(a(a(b(x1))))) b(a(a(b(a(a(a(b(x1)))))))) -> b(b(a(a(a(b(a(a(b(a(a(b(x1)))))))))))) b(a(a(a(b(a(a(a(b(x1))))))))) -> b(x1) b(b(b(a(a(a(b(x1))))))) -> b(a(a(a(b(b(b(x1))))))) b(b(a(a(b(x1))))) -> b(x1) b(a(a(b(b(x1))))) -> b(x1) b(a(b(a(a(a(b(x1))))))) -> b(x1) b(a(a(a(b(a(b(x1))))))) -> 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: B(a(b(x1))) -> B(a(a(a(b(x1))))) B(a(a(b(a(a(a(b(x1)))))))) -> B(b(a(a(a(b(a(a(b(a(a(b(x1)))))))))))) B(a(a(b(a(a(a(b(x1)))))))) -> B(a(a(a(b(a(a(b(a(a(b(x1))))))))))) B(a(a(b(a(a(a(b(x1)))))))) -> B(a(a(b(a(a(b(x1))))))) B(a(a(b(a(a(a(b(x1)))))))) -> B(a(a(b(x1)))) B(b(b(a(a(a(b(x1))))))) -> B(a(a(a(b(b(b(x1))))))) B(b(b(a(a(a(b(x1))))))) -> B(b(b(x1))) B(b(b(a(a(a(b(x1))))))) -> B(b(x1)) The TRS R consists of the following rules: b(a(b(x1))) -> b(a(a(a(b(x1))))) b(a(a(b(a(a(a(b(x1)))))))) -> b(b(a(a(a(b(a(a(b(a(a(b(x1)))))))))))) b(a(a(a(b(a(a(a(b(x1))))))))) -> b(x1) b(b(b(a(a(a(b(x1))))))) -> b(a(a(a(b(b(b(x1))))))) b(b(a(a(b(x1))))) -> b(x1) b(a(a(b(b(x1))))) -> b(x1) b(a(b(a(a(a(b(x1))))))) -> b(x1) b(a(a(a(b(a(b(x1))))))) -> b(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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all output return to SRS Standard