details

property	value
status	complete
benchmark	quadruple2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n004.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.93178915977 seconds
cpu usage	18.695282465
max memory	7.51767552E8

stage attributes

key	value
output-size	17370
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) MRRProof [EQUIVALENT, 10 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) SemLabProof [SOUND, 104 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 9 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) 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: P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) P(p(b(a(x0)), x1), p(x2, x3)) -> P(b(x2), a(a(b(x1)))) P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(b(x2), a(a(b(x1)))), p(x3, x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(b(x2), a(a(b(x1)))), p(x3, x0)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) 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: P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, x0) popout

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