details

property	value
status	complete
benchmark	11.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n177.star.cs.uiowa.edu
space	Various_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.31482 seconds
cpu usage	4.09946
user time	3.92097
system time	0.178486
max virtual memory	1.8345424E7
max residence set size	293324.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) AAECC Innermost [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QReductionProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) TransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) DependencyGraphProof [EQUIVALENT, 0 ms] (20) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y Q is empty. ---------------------------------------- (1) AAECC Innermost (EQUIVALENT) We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is h(0) -> 0 h(g(x, y)) -> y The TRS R 2 is f(0, 1, x) -> f(h(x), h(x), x) The signature Sigma is {f_3} ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y The set Q consists of the following terms: f(0, 1, x0) h(0) h(g(x0, x1)) ---------------------------------------- (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(0, 1, x) -> F(h(x), h(x), x) F(0, 1, x) -> H(x) The TRS R consists of the following rules: f(0, 1, x) -> f(h(x), h(x), x) h(0) -> 0 h(g(x, y)) -> y The set Q consists of the following terms: f(0, 1, x0) h(0) h(g(x0, x1)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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actions

all output return to TRS Standard