details

property	value
status	complete
benchmark	07.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n056.star.cs.uiowa.edu
space	Bouchare_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.03968310356 seconds
cpu usage	16.737475568
max memory	1.262247936E9

stage attributes

key	value
output-size	8171
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) RootLabelingProof [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 6 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 52 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 1 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(x1)) -> b(b(x1)) b(a(b(x1))) -> b(a(a(x1))) a(a(a(x1))) -> a(b(b(x1))) Q is empty. ---------------------------------------- (1) RootLabelingProof (EQUIVALENT) We used plain root labeling [ROOTLAB] with the following heuristic: LabelAll: All function symbols get labeled As Q is empty the root labeling was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(b_{b_1}(x1)) b_{a_1}(a_{a_1}(x1)) -> b_{b_1}(b_{a_1}(x1)) b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{b_1}(x1))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(x1))) a_{a_1}(a_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) a_{a_1}(a_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(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: B_{A_1}(a_{a_1}(x1)) -> B_{A_1}(x1) B_{A_1}(a_{b_1}(b_{b_1}(x1))) -> B_{A_1}(a_{a_1}(a_{b_1}(x1))) B_{A_1}(a_{b_1}(b_{b_1}(x1))) -> A_{A_1}(a_{b_1}(x1)) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> B_{A_1}(a_{a_1}(a_{a_1}(x1))) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(a_{a_1}(x1)) B_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1) A_{A_1}(a_{a_1}(a_{a_1}(x1))) -> B_{A_1}(x1) The TRS R consists of the following rules: b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(b_{b_1}(x1)) b_{a_1}(a_{a_1}(x1)) -> b_{b_1}(b_{a_1}(x1)) b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{a_1}(a_{b_1}(x1))) b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{a_1}(a_{a_1}(x1))) a_{a_1}(a_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(x1))) a_{a_1}(a_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) popout

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