details

property	value
status	complete
benchmark	random-246.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n155.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.81488 seconds
cpu usage	18.5971
user time	17.6972
system time	0.899859
max virtual memory	1.03242676E8
max residence set size	2696600.0

stage attributes

key	value
starexec-result	YES

output

18.03/5.65 YES 18.20/5.75 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 18.20/5.75 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 18.20/5.75 18.20/5.75 18.20/5.75 Termination w.r.t. Q of the given QTRS could be proven: 18.20/5.75 18.20/5.75 (0) QTRS 18.20/5.75 (1) DependencyPairsProof [EQUIVALENT, 3 ms] 18.20/5.75 (2) QDP 18.20/5.75 (3) DependencyGraphProof [EQUIVALENT, 2 ms] 18.20/5.75 (4) AND 18.20/5.75 (5) QDP 18.20/5.75 (6) QDPSizeChangeProof [EQUIVALENT, 0 ms] 18.20/5.75 (7) YES 18.20/5.75 (8) QDP 18.20/5.75 (9) QDPOrderProof [EQUIVALENT, 54 ms] 18.20/5.75 (10) QDP 18.20/5.75 (11) PisEmptyProof [EQUIVALENT, 0 ms] 18.20/5.75 (12) YES 18.20/5.75 18.20/5.75 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (0) 18.20/5.75 Obligation: 18.20/5.75 Q restricted rewrite system: 18.20/5.75 The TRS R consists of the following rules: 18.20/5.75 18.20/5.75 b(a(a(a(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(b(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 18.20/5.75 Q is empty. 18.20/5.75 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (1) DependencyPairsProof (EQUIVALENT) 18.20/5.75 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (2) 18.20/5.75 Obligation: 18.20/5.75 Q DP problem: 18.20/5.75 The TRS P consists of the following rules: 18.20/5.75 18.20/5.75 B(a(a(a(x1)))) -> B(b(a(a(x1)))) 18.20/5.75 B(a(a(a(x1)))) -> B(a(a(x1))) 18.20/5.75 A(b(b(b(x1)))) -> B(b(b(a(x1)))) 18.20/5.75 A(b(b(b(x1)))) -> B(b(a(x1))) 18.20/5.75 A(b(b(b(x1)))) -> B(a(x1)) 18.20/5.75 A(b(b(b(x1)))) -> A(x1) 18.20/5.75 A(b(b(b(x1)))) -> B(b(a(a(x1)))) 18.20/5.75 A(b(b(b(x1)))) -> B(a(a(x1))) 18.20/5.75 A(b(b(b(x1)))) -> A(a(x1)) 18.20/5.75 18.20/5.75 The TRS R consists of the following rules: 18.20/5.75 18.20/5.75 b(a(a(a(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(b(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 18.20/5.75 Q is empty. 18.20/5.75 We have to consider all minimal (P,Q,R)-chains. 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (3) DependencyGraphProof (EQUIVALENT) 18.20/5.75 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes. 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (4) 18.20/5.75 Complex Obligation (AND) 18.20/5.75 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (5) 18.20/5.75 Obligation: 18.20/5.75 Q DP problem: 18.20/5.75 The TRS P consists of the following rules: 18.20/5.75 18.20/5.75 B(a(a(a(x1)))) -> B(a(a(x1))) 18.20/5.75 18.20/5.75 The TRS R consists of the following rules: 18.20/5.75 18.20/5.75 b(a(a(a(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(b(a(x1)))) 18.20/5.75 a(b(b(b(x1)))) -> b(b(a(a(x1)))) 18.20/5.75 18.20/5.75 Q is empty. 18.20/5.75 We have to consider all minimal (P,Q,R)-chains. 18.20/5.75 ---------------------------------------- 18.20/5.75 18.20/5.75 (6) QDPSizeChangeProof (EQUIVALENT) 18.20/5.75 By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 18.20/5.75 18.20/5.75 From the DPs we obtained the following set of size-change graphs: 18.20/5.75 *B(a(a(a(x1)))) -> B(a(a(x1))) 18.20/5.75 The graph contains the following edges 1 > 1 18.20/5.75 18.20/5.75 popout

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