details

property	value
status	complete
benchmark	04.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n116.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	13.2320821285 seconds
cpu usage	48.5896916
max memory	1.761230848E9

stage attributes

key	value
output-size	15727
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 12 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 376 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 719 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 274 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> a(b(b(b(x1)))) b(b(a(b(x1)))) -> a(a(a(x1))) Q is empty. ---------------------------------------- (1) FlatCCProof (EQUIVALENT) We used flat context closure [ROOTLAB] As Q is empty the flat context closure was sound AND complete. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(x1))) -> a(b(b(b(x1)))) a(b(b(a(b(x1))))) -> a(a(a(a(x1)))) b(b(b(a(b(x1))))) -> b(a(a(a(x1)))) Q is empty. ---------------------------------------- (3) 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. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a_{a_1}(a_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1)))) a_{a_1}(a_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1)))) a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) b_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1)))) b_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1)))) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A_{A_1}(a_{a_1}(a_{a_1}(x1))) -> A_{B_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1)))) A_{A_1}(a_{a_1}(a_{a_1}(x1))) -> B_{B_1}(b_{b_1}(b_{a_1}(x1))) popout

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