details

property	value
status	complete
benchmark	random-387.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n016.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	65.4052 seconds
cpu usage	254.171
user time	249.998
system time	4.17352
max virtual memory	6.3024752E7
max residence set size	7044868.0

stage attributes

key	value
starexec-result	YES

output

253.50/65.30 YES 253.80/65.35 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 253.80/65.35 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 253.80/65.35 253.80/65.35 253.80/65.35 Termination w.r.t. Q of the given QTRS could be proven: 253.80/65.35 253.80/65.35 (0) QTRS 253.80/65.35 (1) QTRS Reverse [EQUIVALENT, 0 ms] 253.80/65.35 (2) QTRS 253.80/65.35 (3) DependencyPairsProof [EQUIVALENT, 2 ms] 253.80/65.35 (4) QDP 253.80/65.35 (5) MRRProof [EQUIVALENT, 96 ms] 253.80/65.35 (6) QDP 253.80/65.35 (7) QDPOrderProof [EQUIVALENT, 3249 ms] 253.80/65.35 (8) QDP 253.80/65.35 (9) PisEmptyProof [EQUIVALENT, 0 ms] 253.80/65.35 (10) YES 253.80/65.35 253.80/65.35 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (0) 253.80/65.35 Obligation: 253.80/65.35 Q restricted rewrite system: 253.80/65.35 The TRS R consists of the following rules: 253.80/65.35 253.80/65.35 b(b(b(a(x1)))) -> b(a(a(b(x1)))) 253.80/65.35 b(a(b(a(x1)))) -> a(a(a(b(x1)))) 253.80/65.35 a(a(b(b(x1)))) -> b(b(b(a(x1)))) 253.80/65.35 253.80/65.35 Q is empty. 253.80/65.35 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (1) QTRS Reverse (EQUIVALENT) 253.80/65.35 We applied the QTRS Reverse Processor [REVERSE]. 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (2) 253.80/65.35 Obligation: 253.80/65.35 Q restricted rewrite system: 253.80/65.35 The TRS R consists of the following rules: 253.80/65.35 253.80/65.35 a(b(b(b(x1)))) -> b(a(a(b(x1)))) 253.80/65.35 a(b(a(b(x1)))) -> b(a(a(a(x1)))) 253.80/65.35 b(b(a(a(x1)))) -> a(b(b(b(x1)))) 253.80/65.35 253.80/65.35 Q is empty. 253.80/65.35 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (3) DependencyPairsProof (EQUIVALENT) 253.80/65.35 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (4) 253.80/65.35 Obligation: 253.80/65.35 Q DP problem: 253.80/65.35 The TRS P consists of the following rules: 253.80/65.35 253.80/65.35 A(b(b(b(x1)))) -> B(a(a(b(x1)))) 253.80/65.35 A(b(b(b(x1)))) -> A(a(b(x1))) 253.80/65.35 A(b(b(b(x1)))) -> A(b(x1)) 253.80/65.35 A(b(a(b(x1)))) -> B(a(a(a(x1)))) 253.80/65.35 A(b(a(b(x1)))) -> A(a(a(x1))) 253.80/65.35 A(b(a(b(x1)))) -> A(a(x1)) 253.80/65.35 A(b(a(b(x1)))) -> A(x1) 253.80/65.35 B(b(a(a(x1)))) -> A(b(b(b(x1)))) 253.80/65.35 B(b(a(a(x1)))) -> B(b(b(x1))) 253.80/65.35 B(b(a(a(x1)))) -> B(b(x1)) 253.80/65.35 B(b(a(a(x1)))) -> B(x1) 253.80/65.35 253.80/65.35 The TRS R consists of the following rules: 253.80/65.35 253.80/65.35 a(b(b(b(x1)))) -> b(a(a(b(x1)))) 253.80/65.35 a(b(a(b(x1)))) -> b(a(a(a(x1)))) 253.80/65.35 b(b(a(a(x1)))) -> a(b(b(b(x1)))) 253.80/65.35 253.80/65.35 Q is empty. 253.80/65.35 We have to consider all minimal (P,Q,R)-chains. 253.80/65.35 ---------------------------------------- 253.80/65.35 253.80/65.35 (5) MRRProof (EQUIVALENT) 253.80/65.35 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. 253.80/65.35 253.80/65.35 Strictly oriented dependency pairs: 253.80/65.35 253.80/65.35 A(b(b(b(x1)))) -> A(a(b(x1))) 253.80/65.35 A(b(b(b(x1)))) -> A(b(x1)) 253.80/65.35 A(b(a(b(x1)))) -> A(a(a(x1))) 253.80/65.35 A(b(a(b(x1)))) -> A(a(x1)) 253.80/65.35 A(b(a(b(x1)))) -> A(x1) 253.80/65.35 B(b(a(a(x1)))) -> B(b(b(x1))) 253.80/65.35 B(b(a(a(x1)))) -> B(b(x1)) 253.80/65.35 B(b(a(a(x1)))) -> B(x1) 253.80/65.35 253.80/65.35 253.80/65.35 Used ordering: Polynomial interpretation [POLO]: 253.80/65.35 popout

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