details

property	value
status	complete
benchmark	random-244.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n172.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.24013 seconds
cpu usage	17.0041
user time	16.2478
system time	0.756203
max virtual memory	1.19877832E8
max residence set size	2291180.0

stage attributes

key	value
starexec-result	YES

output

16.29/5.02 YES 16.57/5.11 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 16.57/5.11 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 16.57/5.11 16.57/5.11 16.57/5.11 Termination w.r.t. Q of the given QTRS could be proven: 16.57/5.11 16.57/5.11 (0) QTRS 16.57/5.11 (1) FlatCCProof [EQUIVALENT, 0 ms] 16.57/5.11 (2) QTRS 16.57/5.11 (3) RootLabelingProof [EQUIVALENT, 0 ms] 16.57/5.11 (4) QTRS 16.57/5.11 (5) DependencyPairsProof [EQUIVALENT, 0 ms] 16.57/5.11 (6) QDP 16.57/5.11 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 16.57/5.11 (8) QDP 16.57/5.11 (9) QDPOrderProof [EQUIVALENT, 165 ms] 16.57/5.11 (10) QDP 16.57/5.11 (11) DependencyGraphProof [EQUIVALENT, 0 ms] 16.57/5.11 (12) TRUE 16.57/5.11 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (0) 16.57/5.11 Obligation: 16.57/5.11 Q restricted rewrite system: 16.57/5.11 The TRS R consists of the following rules: 16.57/5.11 16.57/5.11 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 16.57/5.11 b(b(a(b(x1)))) -> a(b(b(a(x1)))) 16.57/5.11 b(b(a(b(x1)))) -> a(a(a(a(x1)))) 16.57/5.11 16.57/5.11 Q is empty. 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (1) FlatCCProof (EQUIVALENT) 16.57/5.11 We used flat context closure [ROOTLAB] 16.57/5.11 As Q is empty the flat context closure was sound AND complete. 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (2) 16.57/5.11 Obligation: 16.57/5.11 Q restricted rewrite system: 16.57/5.11 The TRS R consists of the following rules: 16.57/5.11 16.57/5.11 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 16.57/5.11 a(b(b(a(b(x1))))) -> a(a(b(b(a(x1))))) 16.57/5.11 b(b(b(a(b(x1))))) -> b(a(b(b(a(x1))))) 16.57/5.11 a(b(b(a(b(x1))))) -> a(a(a(a(a(x1))))) 16.57/5.11 b(b(b(a(b(x1))))) -> b(a(a(a(a(x1))))) 16.57/5.11 16.57/5.11 Q is empty. 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (3) RootLabelingProof (EQUIVALENT) 16.57/5.11 We used plain root labeling [ROOTLAB] with the following heuristic: 16.57/5.11 LabelAll: All function symbols get labeled 16.57/5.11 16.57/5.11 As Q is empty the root labeling was sound AND complete. 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (4) 16.57/5.11 Obligation: 16.57/5.11 Q restricted rewrite system: 16.57/5.11 The TRS R consists of the following rules: 16.57/5.11 16.57/5.11 a_{a_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) 16.57/5.11 a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) 16.57/5.11 a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1))))) 16.57/5.11 a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1))))) 16.57/5.11 b_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1))))) 16.57/5.11 b_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1))))) 16.57/5.11 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}(a_{a_1}(x1))))) 16.57/5.11 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_{a_1}(a_{b_1}(x1))))) 16.57/5.11 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}(a_{a_1}(x1))))) 16.57/5.11 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_{a_1}(a_{b_1}(x1))))) 16.57/5.11 16.57/5.11 Q is empty. 16.57/5.11 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (5) DependencyPairsProof (EQUIVALENT) 16.57/5.11 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 16.57/5.11 ---------------------------------------- 16.57/5.11 16.57/5.11 (6) 16.57/5.11 Obligation: 16.57/5.11 Q DP problem: 16.57/5.11 The TRS P consists of the following rules: 16.57/5.11 16.57/5.11 A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) 16.57/5.11 A_{A_1}(a_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> A_{B_1}(b_{a_1}(x1)) 16.57/5.11 A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) 16.57/5.11 A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> A_{B_1}(b_{b_1}(x1)) 16.57/5.11 A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(x1) popout

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all output return to SRS_Standard 2019-03-29 03.29