details

property	value
status	complete
benchmark	z103.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n091.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.3902349472 seconds
cpu usage	13.30539687
max memory	1.729363968E9

stage attributes

key	value
output-size	4410
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDPOrderProof [EQUIVALENT, 14 ms] (6) QDP (7) MRRProof [EQUIVALENT, 10 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) a(x1) -> b(b(b(x1))) b(d(b(x1))) -> a(c(x1)) c(x1) -> d(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(d(x1)) -> B(x1) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) B(d(b(x1))) -> A(c(x1)) B(d(b(x1))) -> C(x1) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) a(x1) -> b(b(b(x1))) b(d(b(x1))) -> a(c(x1)) c(x1) -> d(x1) 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: B(d(b(x1))) -> A(c(x1)) A(d(x1)) -> B(x1) A(x1) -> B(b(b(x1))) A(x1) -> B(b(x1)) A(x1) -> B(x1) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) a(x1) -> b(b(b(x1))) b(d(b(x1))) -> a(c(x1)) c(x1) -> d(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]. The following pairs can be oriented strictly and are deleted. popout

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all output return to SRS Stand 10685