details

property	value
status	complete
benchmark	x03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Secret_07_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	8.19323 seconds
cpu usage	28.3569
user time	27.2707
system time	1.08614
max virtual memory	8.3140228E7
max residence set size	3155884.0

stage attributes

key	value
starexec-result	YES

output

27.58/8.00 YES 28.01/8.03 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 28.01/8.03 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 28.01/8.03 28.01/8.03 28.01/8.03 Termination w.r.t. Q of the given QTRS could be proven: 28.01/8.03 28.01/8.03 (0) QTRS 28.01/8.03 (1) QTRSRRRProof [EQUIVALENT, 58 ms] 28.01/8.03 (2) QTRS 28.01/8.03 (3) DependencyPairsProof [EQUIVALENT, 8 ms] 28.01/8.03 (4) QDP 28.01/8.03 (5) MRRProof [EQUIVALENT, 43 ms] 28.01/8.03 (6) QDP 28.01/8.03 (7) QDPOrderProof [EQUIVALENT, 31 ms] 28.01/8.03 (8) QDP 28.01/8.03 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 28.01/8.03 (10) TRUE 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (0) 28.01/8.03 Obligation: 28.01/8.03 Q restricted rewrite system: 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 a(b(c(x1))) -> b(x1) 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (1) QTRSRRRProof (EQUIVALENT) 28.01/8.03 Used ordering: 28.01/8.03 Polynomial interpretation [POLO]: 28.01/8.03 28.01/8.03 POL(a(x_1)) = 3 + x_1 28.01/8.03 POL(b(x_1)) = 1 + x_1 28.01/8.03 POL(c(x_1)) = 1 + x_1 28.01/8.03 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 28.01/8.03 28.01/8.03 a(b(c(x1))) -> b(x1) 28.01/8.03 28.01/8.03 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (2) 28.01/8.03 Obligation: 28.01/8.03 Q restricted rewrite system: 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (3) DependencyPairsProof (EQUIVALENT) 28.01/8.03 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (4) 28.01/8.03 Obligation: 28.01/8.03 Q DP problem: 28.01/8.03 The TRS P consists of the following rules: 28.01/8.03 28.01/8.03 C(b(b(x1))) -> A(x1) 28.01/8.03 A(a(x1)) -> C(b(a(c(x1)))) 28.01/8.03 A(a(x1)) -> A(c(x1)) 28.01/8.03 A(a(x1)) -> C(x1) 28.01/8.03 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 We have to consider all minimal (P,Q,R)-chains. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (5) MRRProof (EQUIVALENT) 28.01/8.03 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. 28.01/8.03 28.01/8.03 Strictly oriented dependency pairs: 28.01/8.03 28.01/8.03 A(a(x1)) -> A(c(x1)) 28.01/8.03 A(a(x1)) -> C(x1) 28.01/8.03 28.01/8.03 28.01/8.03 Used ordering: Polynomial interpretation [POLO]: 28.01/8.03 popout

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all output return to SRS_Standard 2019-03-29 03.29