details

property	value
status	complete
benchmark	random-8.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.10459 seconds
cpu usage	31.0897
user time	29.5544
system time	1.53534
max virtual memory	1.02909856E8
max residence set size	4022180.0

stage attributes

key	value
starexec-result	YES

output

29.93/8.85 YES 30.69/8.99 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 30.69/8.99 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 30.69/8.99 30.69/8.99 30.69/8.99 Termination w.r.t. Q of the given QTRS could be proven: 30.69/8.99 30.69/8.99 (0) QTRS 30.69/8.99 (1) QTRS Reverse [EQUIVALENT, 0 ms] 30.69/8.99 (2) QTRS 30.69/8.99 (3) DependencyPairsProof [EQUIVALENT, 23 ms] 30.69/8.99 (4) QDP 30.69/8.99 (5) MRRProof [EQUIVALENT, 89 ms] 30.69/8.99 (6) QDP 30.69/8.99 (7) QDPOrderProof [EQUIVALENT, 546 ms] 30.69/8.99 (8) QDP 30.69/8.99 (9) QDPOrderProof [EQUIVALENT, 62 ms] 30.69/8.99 (10) QDP 30.69/8.99 (11) DependencyGraphProof [EQUIVALENT, 0 ms] 30.69/8.99 (12) TRUE 30.69/8.99 30.69/8.99 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (0) 30.69/8.99 Obligation: 30.69/8.99 Q restricted rewrite system: 30.69/8.99 The TRS R consists of the following rules: 30.69/8.99 30.69/8.99 b(a(a(a(x1)))) -> b(b(a(b(x1)))) 30.69/8.99 b(b(b(a(x1)))) -> b(b(a(a(x1)))) 30.69/8.99 a(a(b(b(x1)))) -> a(a(a(a(x1)))) 30.69/8.99 30.69/8.99 Q is empty. 30.69/8.99 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (1) QTRS Reverse (EQUIVALENT) 30.69/8.99 We applied the QTRS Reverse Processor [REVERSE]. 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (2) 30.69/8.99 Obligation: 30.69/8.99 Q restricted rewrite system: 30.69/8.99 The TRS R consists of the following rules: 30.69/8.99 30.69/8.99 a(a(a(b(x1)))) -> b(a(b(b(x1)))) 30.69/8.99 a(b(b(b(x1)))) -> a(a(b(b(x1)))) 30.69/8.99 b(b(a(a(x1)))) -> a(a(a(a(x1)))) 30.69/8.99 30.69/8.99 Q is empty. 30.69/8.99 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (3) DependencyPairsProof (EQUIVALENT) 30.69/8.99 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (4) 30.69/8.99 Obligation: 30.69/8.99 Q DP problem: 30.69/8.99 The TRS P consists of the following rules: 30.69/8.99 30.69/8.99 A(a(a(b(x1)))) -> B(a(b(b(x1)))) 30.69/8.99 A(a(a(b(x1)))) -> A(b(b(x1))) 30.69/8.99 A(a(a(b(x1)))) -> B(b(x1)) 30.69/8.99 A(b(b(b(x1)))) -> A(a(b(b(x1)))) 30.69/8.99 A(b(b(b(x1)))) -> A(b(b(x1))) 30.69/8.99 B(b(a(a(x1)))) -> A(a(a(a(x1)))) 30.69/8.99 B(b(a(a(x1)))) -> A(a(a(x1))) 30.69/8.99 30.69/8.99 The TRS R consists of the following rules: 30.69/8.99 30.69/8.99 a(a(a(b(x1)))) -> b(a(b(b(x1)))) 30.69/8.99 a(b(b(b(x1)))) -> a(a(b(b(x1)))) 30.69/8.99 b(b(a(a(x1)))) -> a(a(a(a(x1)))) 30.69/8.99 30.69/8.99 Q is empty. 30.69/8.99 We have to consider all minimal (P,Q,R)-chains. 30.69/8.99 ---------------------------------------- 30.69/8.99 30.69/8.99 (5) MRRProof (EQUIVALENT) 30.69/8.99 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. 30.69/8.99 30.69/8.99 Strictly oriented dependency pairs: 30.69/8.99 30.69/8.99 A(a(a(b(x1)))) -> A(b(b(x1))) 30.69/8.99 A(a(a(b(x1)))) -> B(b(x1)) 30.69/8.99 A(b(b(b(x1)))) -> A(b(b(x1))) 30.69/8.99 B(b(a(a(x1)))) -> A(a(a(x1))) 30.69/8.99 30.69/8.99 30.69/8.99 Used ordering: Polynomial interpretation [POLO]: 30.69/8.99 30.69/8.99 POL(A(x_1)) = 2*x_1 30.69/8.99 POL(B(x_1)) = 2*x_1 30.69/8.99 POL(a(x_1)) = 2 + x_1 30.69/8.99 POL(b(x_1)) = 2 + x_1 30.69/8.99 30.69/8.99 popout

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all output return to SRS_Standard 2019-03-29 03.29