details

property	value
status	complete
benchmark	n003.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n189.star.cs.uiowa.edu
space	HirokawaMiddeldorp_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.14806 seconds
cpu usage	5.26244
user time	4.9937
system time	0.268746
max virtual memory	2.0273952E7
max residence set size	513828.0

stage attributes

key	value
starexec-result	NO

output

NO 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QReductionProof [EQUIVALENT, 0 ms] (8) QDP (9) NonTerminationLoopProof [COMPLETE, 0 ms] (10) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x) -> f(a) Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none The TRS R 2 is f(x) -> f(a) The signature Sigma is {f_1} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x) -> f(a) The set Q consists of the following terms: f(x0) ---------------------------------------- (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: F(x) -> F(a) The TRS R consists of the following rules: f(x) -> f(a) The set Q consists of the following terms: f(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: F(x) -> F(a) R is empty. The set Q consists of the following terms: f(x0) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. popout

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job log

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actions

all output return to TRS Standard