details

property	value
status	complete
benchmark	random-284.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n123.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.63972210884 seconds
cpu usage	26.77218272
max memory	3.870101504E9

stage attributes

key	value
output-size	5688
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 40 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 6 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 33 ms] (6) QDP (7) MRRProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 65 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(x1)))) -> b(b(a(a(x1)))) b(a(b(a(x1)))) -> a(b(a(b(x1)))) b(a(b(b(x1)))) -> a(b(a(a(x1)))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = x_1 POL(b(x_1)) = 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(b(b(x1)))) -> a(b(a(a(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(x1)))) -> b(b(a(a(x1)))) b(a(b(a(x1)))) -> a(b(a(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(b(a(b(x1)))) -> B(b(a(a(x1)))) A(b(a(b(x1)))) -> B(a(a(x1))) A(b(a(b(x1)))) -> A(a(x1)) A(b(a(b(x1)))) -> A(x1) B(a(b(a(x1)))) -> A(b(a(b(x1)))) B(a(b(a(x1)))) -> B(a(b(x1))) B(a(b(a(x1)))) -> A(b(x1)) B(a(b(a(x1)))) -> B(x1) The TRS R consists of the following rules: a(b(a(b(x1)))) -> b(b(a(a(x1)))) b(a(b(a(x1)))) -> a(b(a(b(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. popout

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