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	n119.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	10.1134541035 seconds
cpu usage	19.956073597
max memory	8.77207552E8

stage attributes

key	value
output-size	17367
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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, 24 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, 117 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) QDP (17) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 0 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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actions

all output return to TRS Standard