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	n139.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	12.9673 seconds
cpu usage	47.269
user time	44.9511
system time	2.31787
max virtual memory	6.043898E7
max residence set size	4634080.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) FlatCCProof [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 31 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 379 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 735 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 279 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))) A_{A_1}(a_{a_1}(a_{a_1}(x1))) -> 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_{A_1}(a_{a_1}(a_{b_1}(x1))) -> B_{B_1}(b_{b_1}(b_{b_1}(x1))) A_{A_1}(a_{a_1}(a_{b_1}(x1))) -> B_{B_1}(b_{b_1}(x1)) A_{A_1}(a_{a_1}(a_{b_1}(x1))) -> 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)))) popout

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