details

property	value
status	complete
benchmark	z126.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.00203 seconds
cpu usage	5.18438
user time	4.95998
system time	0.224394
max virtual memory	1.9275272E7
max residence set size	528000.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) RootLabelingProof [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(x1))) -> a(b(b(a(x1)))) b(b(b(x1))) -> b(b(x1)) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(x1))) -> a(b(b(a(x1)))) b(b(b(x1))) -> b(b(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_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) b_{b_1}(b_{b_1}(b_{a_1}(x1))) -> b_{b_1}(b_{a_1}(x1)) b_{b_1}(b_{b_1}(b_{b_1}(x1))) -> b_{b_1}(b_{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_{B_1}(b_{a_1}(a_{a_1}(x1))) -> A_{B_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) A_{B_1}(b_{a_1}(a_{a_1}(x1))) -> B_{B_1}(b_{a_1}(a_{a_1}(x1))) A_{B_1}(b_{a_1}(a_{b_1}(x1))) -> A_{B_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) A_{B_1}(b_{a_1}(a_{b_1}(x1))) -> B_{B_1}(b_{a_1}(a_{b_1}(x1))) The TRS R consists of the following rules: a_{b_1}(b_{a_1}(a_{a_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) a_{b_1}(b_{a_1}(a_{b_1}(x1))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) b_{b_1}(b_{b_1}(b_{a_1}(x1))) -> b_{b_1}(b_{a_1}(x1)) b_{b_1}(b_{b_1}(b_{b_1}(x1))) -> b_{b_1}(b_{b_1}(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes. ---------------------------------------- (8) popout

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