details

property	value
status	complete
benchmark	random-113.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n042.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	6.50264 seconds
cpu usage	21.7421
user time	20.7735
system time	0.968543
max virtual memory	1.2225664E8
max residence set size	2825008.0

stage attributes

key	value
starexec-result	YES

output

21.38/6.39 YES 21.38/6.40 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 21.38/6.40 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 21.38/6.40 21.38/6.40 21.38/6.40 Termination w.r.t. Q of the given QTRS could be proven: 21.38/6.40 21.38/6.40 (0) QTRS 21.38/6.40 (1) DependencyPairsProof [EQUIVALENT, 0 ms] 21.38/6.40 (2) QDP 21.38/6.40 (3) QDPOrderProof [EQUIVALENT, 27 ms] 21.38/6.40 (4) QDP 21.38/6.40 (5) QDPOrderProof [EQUIVALENT, 127 ms] 21.38/6.40 (6) QDP 21.38/6.40 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 21.38/6.40 (8) TRUE 21.38/6.40 21.38/6.40 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (0) 21.38/6.40 Obligation: 21.38/6.40 Q restricted rewrite system: 21.38/6.40 The TRS R consists of the following rules: 21.38/6.40 21.38/6.40 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 21.38/6.40 a(b(a(a(x1)))) -> a(a(a(a(x1)))) 21.38/6.40 a(a(a(b(x1)))) -> b(b(b(b(x1)))) 21.38/6.40 21.38/6.40 Q is empty. 21.38/6.40 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (1) DependencyPairsProof (EQUIVALENT) 21.38/6.40 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (2) 21.38/6.40 Obligation: 21.38/6.40 Q DP problem: 21.38/6.40 The TRS P consists of the following rules: 21.38/6.40 21.38/6.40 A(a(b(a(x1)))) -> A(b(a(b(x1)))) 21.38/6.40 A(a(b(a(x1)))) -> A(b(x1)) 21.38/6.40 A(b(a(a(x1)))) -> A(a(a(a(x1)))) 21.38/6.40 A(b(a(a(x1)))) -> A(a(a(x1))) 21.38/6.40 21.38/6.40 The TRS R consists of the following rules: 21.38/6.40 21.38/6.40 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 21.38/6.40 a(b(a(a(x1)))) -> a(a(a(a(x1)))) 21.38/6.40 a(a(a(b(x1)))) -> b(b(b(b(x1)))) 21.38/6.40 21.38/6.40 Q is empty. 21.38/6.40 We have to consider all minimal (P,Q,R)-chains. 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (3) QDPOrderProof (EQUIVALENT) 21.38/6.40 We use the reduction pair processor [LPAR04,JAR06]. 21.38/6.40 21.38/6.40 21.38/6.40 The following pairs can be oriented strictly and are deleted. 21.38/6.40 21.38/6.40 A(a(b(a(x1)))) -> A(b(x1)) 21.38/6.40 A(b(a(a(x1)))) -> A(a(a(x1))) 21.38/6.40 The remaining pairs can at least be oriented weakly. 21.38/6.40 Used ordering: Polynomial interpretation [POLO]: 21.38/6.40 21.38/6.40 POL(A(x_1)) = x_1 21.38/6.40 POL(a(x_1)) = 1 + x_1 21.38/6.40 POL(b(x_1)) = 1 + x_1 21.38/6.40 21.38/6.40 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 21.38/6.40 21.38/6.40 a(b(a(a(x1)))) -> a(a(a(a(x1)))) 21.38/6.40 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 21.38/6.40 a(a(a(b(x1)))) -> b(b(b(b(x1)))) 21.38/6.40 21.38/6.40 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (4) 21.38/6.40 Obligation: 21.38/6.40 Q DP problem: 21.38/6.40 The TRS P consists of the following rules: 21.38/6.40 21.38/6.40 A(a(b(a(x1)))) -> A(b(a(b(x1)))) 21.38/6.40 A(b(a(a(x1)))) -> A(a(a(a(x1)))) 21.38/6.40 21.38/6.40 The TRS R consists of the following rules: 21.38/6.40 21.38/6.40 a(a(b(a(x1)))) -> a(b(a(b(x1)))) 21.38/6.40 a(b(a(a(x1)))) -> a(a(a(a(x1)))) 21.38/6.40 a(a(a(b(x1)))) -> b(b(b(b(x1)))) 21.38/6.40 21.38/6.40 Q is empty. 21.38/6.40 We have to consider all minimal (P,Q,R)-chains. 21.38/6.40 ---------------------------------------- 21.38/6.40 21.38/6.40 (5) QDPOrderProof (EQUIVALENT) popout

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