details

property	value
status	complete
benchmark	z117.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n018.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	10.2184448242 seconds
cpu usage	36.62549473
max memory	3.986178048E9

stage attributes

key	value
output-size	7088
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 52 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 43 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 1253 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: d(a(x1)) -> b(d(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> a(d(x1)) b(d(d(x1))) -> c(c(d(d(c(x1))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> d(a(x1)) d(d(b(x1))) -> c(d(d(c(c(x1))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(d(x1)) -> D(b(x1)) A(d(x1)) -> B(x1) B(x1) -> A(a(a(x1))) B(x1) -> A(a(x1)) B(x1) -> A(x1) C(d(c(x1))) -> D(a(x1)) C(d(c(x1))) -> A(x1) D(d(b(x1))) -> C(d(d(c(c(x1))))) D(d(b(x1))) -> D(d(c(c(x1)))) D(d(b(x1))) -> D(c(c(x1))) D(d(b(x1))) -> C(c(x1)) D(d(b(x1))) -> C(x1) The TRS R consists of the following rules: a(d(x1)) -> d(b(x1)) b(x1) -> a(a(a(x1))) c(d(c(x1))) -> d(a(x1)) d(d(b(x1))) -> c(d(d(c(c(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]. popout

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job log

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actions

all output return to SRS Standard