details

property	value
status	complete
benchmark	z18.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.63400578499 seconds
cpu usage	3.737756553
max memory	2.32951808E8

stage attributes

key	value
output-size	4841
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPToSRSProof [SOUND, 0 ms] (6) QTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) QDP (9) MNOCProof [EQUIVALENT, 0 ms] (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QReductionProof [EQUIVALENT, 0 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 0 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(x, a(b(y))) -> a(f(a(b(x)), y)) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) 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(x, a(b(y))) -> F(a(b(x)), y) F(a(x), y) -> F(x, a(y)) F(b(x), y) -> F(x, b(y)) The TRS R consists of the following rules: f(x, a(b(y))) -> a(f(a(b(x)), y)) f(a(x), y) -> f(x, a(y)) f(b(x), y) -> f(x, b(y)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) 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. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(x, a(b(y))) -> F(a(b(x)), y) F(a(x), y) -> F(x, a(y)) F(b(x), y) -> F(x, b(y)) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPToSRSProof (SOUND) The finiteness of this DP problem is implied by strong termination of a SRS due to [UNKNOWN]. ---------------------------------------- (6) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x)) -> b(a(x)) popout

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