details

property	value
status	complete
benchmark	quadruple1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	Endrullis_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.22017002106 seconds
cpu usage	4.815251597
max memory	3.12717312E8

stage attributes

key	value
output-size	3872
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 0 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) 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(x3, a(x2)), p(b(a(x1)), b(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(x3, a(x2)), p(b(a(x1)), b(x0))) P(p(b(a(x0)), x1), p(x2, x3)) -> P(x3, a(x2)) P(p(b(a(x0)), x1), p(x2, x3)) -> P(b(a(x1)), b(x0)) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(x3, a(x2)), p(b(a(x1)), b(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 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(x3, a(x2)), p(b(a(x1)), b(x0))) The TRS R consists of the following rules: p(p(b(a(x0)), x1), p(x2, x3)) -> p(p(x3, a(x2)), p(b(a(x1)), b(x0))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: P(p(b(a(x0)), x1), p(x2, x3)) -> P(p(x3, a(x2)), p(b(a(x1)), b(x0))) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. popout

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