details

property	value
status	complete
benchmark	random-425.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n046.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.09722 seconds
cpu usage	24.1978
user time	22.971
system time	1.22674
max virtual memory	4.1916792E7
max residence set size	3593136.0

stage attributes

key	value
starexec-result	YES

output

23.74/7.04 YES 23.74/7.05 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 23.74/7.05 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 23.74/7.05 23.74/7.05 23.74/7.05 Termination w.r.t. Q of the given QTRS could be proven: 23.74/7.05 23.74/7.05 (0) QTRS 23.74/7.05 (1) DependencyPairsProof [EQUIVALENT, 27 ms] 23.74/7.05 (2) QDP 23.74/7.05 (3) MRRProof [EQUIVALENT, 71 ms] 23.74/7.05 (4) QDP 23.74/7.05 (5) MRRProof [EQUIVALENT, 13 ms] 23.74/7.05 (6) QDP 23.74/7.05 (7) PisEmptyProof [EQUIVALENT, 0 ms] 23.74/7.05 (8) YES 23.74/7.05 23.74/7.05 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (0) 23.74/7.05 Obligation: 23.74/7.05 Q restricted rewrite system: 23.74/7.05 The TRS R consists of the following rules: 23.74/7.05 23.74/7.05 b(a(a(b(x1)))) -> b(b(b(a(x1)))) 23.74/7.05 b(b(b(b(x1)))) -> a(b(b(b(x1)))) 23.74/7.05 b(a(b(b(x1)))) -> b(a(b(a(x1)))) 23.74/7.05 23.74/7.05 Q is empty. 23.74/7.05 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (1) DependencyPairsProof (EQUIVALENT) 23.74/7.05 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (2) 23.74/7.05 Obligation: 23.74/7.05 Q DP problem: 23.74/7.05 The TRS P consists of the following rules: 23.74/7.05 23.74/7.05 B(a(a(b(x1)))) -> B(b(b(a(x1)))) 23.74/7.05 B(a(a(b(x1)))) -> B(b(a(x1))) 23.74/7.05 B(a(a(b(x1)))) -> B(a(x1)) 23.74/7.05 B(a(b(b(x1)))) -> B(a(b(a(x1)))) 23.74/7.05 B(a(b(b(x1)))) -> B(a(x1)) 23.74/7.05 23.74/7.05 The TRS R consists of the following rules: 23.74/7.05 23.74/7.05 b(a(a(b(x1)))) -> b(b(b(a(x1)))) 23.74/7.05 b(b(b(b(x1)))) -> a(b(b(b(x1)))) 23.74/7.05 b(a(b(b(x1)))) -> b(a(b(a(x1)))) 23.74/7.05 23.74/7.05 Q is empty. 23.74/7.05 We have to consider all minimal (P,Q,R)-chains. 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (3) MRRProof (EQUIVALENT) 23.74/7.05 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 23.74/7.05 23.74/7.05 Strictly oriented dependency pairs: 23.74/7.05 23.74/7.05 B(a(a(b(x1)))) -> B(b(a(x1))) 23.74/7.05 B(a(a(b(x1)))) -> B(a(x1)) 23.74/7.05 B(a(b(b(x1)))) -> B(a(x1)) 23.74/7.05 23.74/7.05 23.74/7.05 Used ordering: Polynomial interpretation [POLO]: 23.74/7.05 23.74/7.05 POL(B(x_1)) = 2*x_1 23.74/7.05 POL(a(x_1)) = 2 + x_1 23.74/7.05 POL(b(x_1)) = 2 + x_1 23.74/7.05 23.74/7.05 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (4) 23.74/7.05 Obligation: 23.74/7.05 Q DP problem: 23.74/7.05 The TRS P consists of the following rules: 23.74/7.05 23.74/7.05 B(a(a(b(x1)))) -> B(b(b(a(x1)))) 23.74/7.05 B(a(b(b(x1)))) -> B(a(b(a(x1)))) 23.74/7.05 23.74/7.05 The TRS R consists of the following rules: 23.74/7.05 23.74/7.05 b(a(a(b(x1)))) -> b(b(b(a(x1)))) 23.74/7.05 b(b(b(b(x1)))) -> a(b(b(b(x1)))) 23.74/7.05 b(a(b(b(x1)))) -> b(a(b(a(x1)))) 23.74/7.05 23.74/7.05 Q is empty. 23.74/7.05 We have to consider all minimal (P,Q,R)-chains. 23.74/7.05 ---------------------------------------- 23.74/7.05 23.74/7.05 (5) MRRProof (EQUIVALENT) 23.74/7.05 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 23.74/7.05 23.74/7.05 Strictly oriented dependency pairs: 23.74/7.05 popout

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all output return to SRS_Standard 2019-03-29 03.29