details

property	value
status	complete
benchmark	14.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n088.star.cs.uiowa.edu
space	Gebhardt_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.96879 seconds
cpu usage	35.9656
user time	34.4545
system time	1.51103
max virtual memory	6.2815252E7
max residence set size	4206012.0

stage attributes

key	value
starexec-result	YES

output

35.54/9.83 YES 35.70/9.87 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 35.70/9.87 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 35.70/9.87 35.70/9.87 35.70/9.87 Termination w.r.t. Q of the given QTRS could be proven: 35.70/9.87 35.70/9.87 (0) QTRS 35.70/9.87 (1) DependencyPairsProof [EQUIVALENT, 15 ms] 35.70/9.87 (2) QDP 35.70/9.87 (3) MRRProof [EQUIVALENT, 68 ms] 35.70/9.87 (4) QDP 35.70/9.87 (5) QDPOrderProof [EQUIVALENT, 107 ms] 35.70/9.87 (6) QDP 35.70/9.87 (7) PisEmptyProof [EQUIVALENT, 0 ms] 35.70/9.87 (8) YES 35.70/9.87 35.70/9.87 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (0) 35.70/9.87 Obligation: 35.70/9.87 Q restricted rewrite system: 35.70/9.87 The TRS R consists of the following rules: 35.70/9.87 35.70/9.87 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 35.70/9.87 1(1(0(1(x1)))) -> 0(0(0(0(x1)))) 35.70/9.87 35.70/9.87 Q is empty. 35.70/9.87 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (1) DependencyPairsProof (EQUIVALENT) 35.70/9.87 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (2) 35.70/9.87 Obligation: 35.70/9.87 Q DP problem: 35.70/9.87 The TRS P consists of the following rules: 35.70/9.87 35.70/9.87 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(x1) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1)))) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(0(x1))) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(x1)) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(x1) 35.70/9.87 35.70/9.87 The TRS R consists of the following rules: 35.70/9.87 35.70/9.87 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 35.70/9.87 1(1(0(1(x1)))) -> 0(0(0(0(x1)))) 35.70/9.87 35.70/9.87 Q is empty. 35.70/9.87 We have to consider all minimal (P,Q,R)-chains. 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (3) MRRProof (EQUIVALENT) 35.70/9.87 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. 35.70/9.87 35.70/9.87 Strictly oriented dependency pairs: 35.70/9.87 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(1(1(x1))) 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(1(x1)) 35.70/9.87 0^1(0(0(0(x1)))) -> 1^1(x1) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1)))) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(0(x1))) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(0(x1)) 35.70/9.87 1^1(1(0(1(x1)))) -> 0^1(x1) 35.70/9.87 35.70/9.87 35.70/9.87 Used ordering: Polynomial interpretation [POLO]: 35.70/9.87 35.70/9.87 POL(0(x_1)) = 2 + x_1 35.70/9.87 POL(0^1(x_1)) = 1 + x_1 35.70/9.87 POL(1(x_1)) = 2 + x_1 35.70/9.87 POL(1^1(x_1)) = 2 + x_1 35.70/9.87 35.70/9.87 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (4) 35.70/9.87 Obligation: 35.70/9.87 Q DP problem: 35.70/9.87 The TRS P consists of the following rules: 35.70/9.87 35.70/9.87 0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1)))) 35.70/9.87 35.70/9.87 The TRS R consists of the following rules: 35.70/9.87 35.70/9.87 0(0(0(0(x1)))) -> 0(1(1(1(x1)))) 35.70/9.87 1(1(0(1(x1)))) -> 0(0(0(0(x1)))) 35.70/9.87 35.70/9.87 Q is empty. 35.70/9.87 We have to consider all minimal (P,Q,R)-chains. 35.70/9.87 ---------------------------------------- 35.70/9.87 35.70/9.87 (5) QDPOrderProof (EQUIVALENT) popout

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