details

property	value
status	complete
benchmark	teparla2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n073.star.cs.uiowa.edu
space	Secret_05_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.8056 seconds
cpu usage	7.35968
user time	7.06245
system time	0.297228
max virtual memory	3.7131304E7
max residence set size	503432.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 12 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) SemLabProof [SOUND, 91 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) MRRProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (16) QDP (17) QDPOrderProof [EQUIVALENT, 5 ms] (18) QDP (19) PisEmptyProof [EQUIVALENT, 0 ms] (20) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(y, f(x, f(a, x))) -> f(f(f(a, x), f(x, a)), f(a, y)) f(x, f(x, y)) -> f(f(f(x, a), a), a) 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: F(y, f(x, f(a, x))) -> F(f(f(a, x), f(x, a)), f(a, y)) F(y, f(x, f(a, x))) -> F(f(a, x), f(x, a)) F(y, f(x, f(a, x))) -> F(x, a) F(y, f(x, f(a, x))) -> F(a, y) F(x, f(x, y)) -> F(f(f(x, a), a), a) F(x, f(x, y)) -> F(f(x, a), a) F(x, f(x, y)) -> F(x, a) The TRS R consists of the following rules: f(y, f(x, f(a, x))) -> f(f(f(a, x), f(x, a)), f(a, y)) f(x, f(x, y)) -> f(f(f(x, a), a), a) 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 5 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(y, f(x, f(a, x))) -> F(a, y) F(y, f(x, f(a, x))) -> F(f(f(a, x), f(x, a)), f(a, y)) The TRS R consists of the following rules: f(y, f(x, f(a, x))) -> f(f(f(a, x), f(x, a)), f(a, y)) f(x, f(x, y)) -> f(f(f(x, a), a), a) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) SemLabProof (SOUND) We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1. a: 1 f: 0 F: 0 By semantic labelling [SEMLAB] we obtain the following labelled QDP problem. ---------------------------------------- popout

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