details

property	value
status	complete
benchmark	01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.60912203789 seconds
cpu usage	10.828577599
max memory	1.187131392E9

stage attributes

key	value
output-size	5729
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 52 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) UsableRulesProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x1)) -> b(r(x1)) r(a(x1)) -> d(r(x1)) r(x1) -> d(x1) d(a(x1)) -> a(a(d(x1))) d(x1) -> a(x1) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(x1)) -> R(x1) R(a(x1)) -> D(r(x1)) R(a(x1)) -> R(x1) R(x1) -> D(x1) D(a(x1)) -> A(a(d(x1))) D(a(x1)) -> A(d(x1)) D(a(x1)) -> D(x1) D(x1) -> A(x1) The TRS R consists of the following rules: a(b(x1)) -> b(r(x1)) r(a(x1)) -> d(r(x1)) r(x1) -> d(x1) d(a(x1)) -> a(a(d(x1))) d(x1) -> a(x1) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. Strictly oriented dependency pairs: A(b(x1)) -> R(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(D(x_1)) = 2*x_1 POL(R(x_1)) = 3*x_1 POL(a(x_1)) = x_1 POL(b(x_1)) = 1 + 3*x_1 POL(d(x_1)) = x_1 POL(r(x_1)) = x_1 ---------------------------------------- popout

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

job log

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actions

all output return to SRS Standard