details

property	value
status	complete
benchmark	random-546.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n173.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.49242 seconds
cpu usage	17.7812
user time	17.0638
system time	0.717398
max virtual memory	3.7240304E7
max residence set size	2055748.0

stage attributes

key	value
starexec-result	YES

output

17.59/5.42 YES 17.59/5.43 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 17.59/5.43 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 17.59/5.43 17.59/5.43 17.59/5.43 Termination w.r.t. Q of the given QTRS could be proven: 17.59/5.43 17.59/5.43 (0) QTRS 17.59/5.43 (1) FlatCCProof [EQUIVALENT, 0 ms] 17.59/5.43 (2) QTRS 17.59/5.43 (3) RootLabelingProof [EQUIVALENT, 1 ms] 17.59/5.43 (4) QTRS 17.59/5.43 (5) DependencyPairsProof [EQUIVALENT, 87 ms] 17.59/5.43 (6) QDP 17.59/5.43 (7) DependencyGraphProof [EQUIVALENT, 7 ms] 17.59/5.43 (8) QDP 17.59/5.43 (9) QDPOrderProof [EQUIVALENT, 76 ms] 17.59/5.43 (10) QDP 17.59/5.43 (11) DependencyGraphProof [EQUIVALENT, 0 ms] 17.59/5.43 (12) TRUE 17.59/5.43 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (0) 17.59/5.43 Obligation: 17.59/5.43 Q restricted rewrite system: 17.59/5.43 The TRS R consists of the following rules: 17.59/5.43 17.59/5.43 b(b(b(b(x1)))) -> a(a(b(b(x1)))) 17.59/5.43 a(a(b(a(x1)))) -> a(a(a(b(x1)))) 17.59/5.43 b(a(a(a(x1)))) -> a(b(a(b(x1)))) 17.59/5.43 17.59/5.43 Q is empty. 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (1) FlatCCProof (EQUIVALENT) 17.59/5.43 We used flat context closure [ROOTLAB] 17.59/5.43 As Q is empty the flat context closure was sound AND complete. 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (2) 17.59/5.43 Obligation: 17.59/5.43 Q restricted rewrite system: 17.59/5.43 The TRS R consists of the following rules: 17.59/5.43 17.59/5.43 a(a(b(a(x1)))) -> a(a(a(b(x1)))) 17.59/5.43 b(b(b(b(b(x1))))) -> b(a(a(b(b(x1))))) 17.59/5.43 a(b(b(b(b(x1))))) -> a(a(a(b(b(x1))))) 17.59/5.43 b(b(a(a(a(x1))))) -> b(a(b(a(b(x1))))) 17.59/5.43 a(b(a(a(a(x1))))) -> a(a(b(a(b(x1))))) 17.59/5.43 17.59/5.43 Q is empty. 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (3) RootLabelingProof (EQUIVALENT) 17.59/5.43 We used plain root labeling [ROOTLAB] with the following heuristic: 17.59/5.43 LabelAll: All function symbols get labeled 17.59/5.43 17.59/5.43 As Q is empty the root labeling was sound AND complete. 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (4) 17.59/5.43 Obligation: 17.59/5.43 Q restricted rewrite system: 17.59/5.43 The TRS R consists of the following rules: 17.59/5.43 17.59/5.43 a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))) 17.59/5.43 a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))) 17.59/5.43 b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) 17.59/5.43 b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) 17.59/5.43 a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))) 17.59/5.43 a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))) 17.59/5.43 b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) 17.59/5.43 b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) 17.59/5.43 a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) 17.59/5.43 a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) 17.59/5.43 17.59/5.43 Q is empty. 17.59/5.43 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (5) DependencyPairsProof (EQUIVALENT) 17.59/5.43 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 17.59/5.43 ---------------------------------------- 17.59/5.43 17.59/5.43 (6) 17.59/5.43 Obligation: 17.59/5.43 Q DP problem: 17.59/5.43 The TRS P consists of the following rules: 17.59/5.43 17.59/5.43 A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{A_1}(a_{a_1}(a_{b_1}(b_{a_1}(x1)))) 17.59/5.43 A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{A_1}(a_{b_1}(b_{a_1}(x1))) 17.59/5.43 A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{B_1}(b_{a_1}(x1)) 17.59/5.43 A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> A_{A_1}(a_{a_1}(a_{b_1}(b_{b_1}(x1)))) 17.59/5.43 A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> A_{A_1}(a_{b_1}(b_{b_1}(x1))) popout

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all output return to SRS_Standard 2019-03-29 03.29