details

property	value
status	complete
benchmark	dup17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n051.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	9.81151 seconds
cpu usage	34.7938
user time	33.2638
system time	1.53003
max virtual memory	6.1587368E7
max residence set size	4214092.0

stage attributes

key	value
starexec-result	YES

output

34.44/9.71 YES 34.44/9.75 proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml 34.44/9.75 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 34.44/9.75 34.44/9.75 34.44/9.75 Termination w.r.t. Q of the given QTRS could be proven: 34.44/9.75 34.44/9.75 (0) QTRS 34.44/9.75 (1) DependencyPairsProof [EQUIVALENT, 51 ms] 34.44/9.75 (2) QDP 34.44/9.75 (3) QDPOrderProof [EQUIVALENT, 175 ms] 34.44/9.75 (4) QDP 34.44/9.75 (5) MRRProof [EQUIVALENT, 109 ms] 34.44/9.75 (6) QDP 34.44/9.75 (7) DependencyGraphProof [EQUIVALENT, 1 ms] 34.44/9.75 (8) QDP 34.44/9.75 (9) MRRProof [EQUIVALENT, 0 ms] 34.44/9.75 (10) QDP 34.44/9.75 (11) MRRProof [EQUIVALENT, 12 ms] 34.44/9.75 (12) QDP 34.44/9.75 (13) PisEmptyProof [EQUIVALENT, 0 ms] 34.44/9.75 (14) YES 34.44/9.75 34.44/9.75 34.44/9.75 ---------------------------------------- 34.44/9.75 34.44/9.75 (0) 34.44/9.75 Obligation: 34.44/9.75 Q restricted rewrite system: 34.44/9.75 The TRS R consists of the following rules: 34.44/9.75 34.44/9.75 a(a(a(a(x1)))) -> b(b(x1)) 34.44/9.75 b(b(a(a(x1)))) -> a(a(b(b(x1)))) 34.44/9.75 b(b(b(b(c(c(x1)))))) -> c(c(a(a(x1)))) 34.44/9.75 b(b(b(b(x1)))) -> a(a(a(a(a(a(x1)))))) 34.44/9.75 c(c(a(a(x1)))) -> b(b(a(a(c(c(x1)))))) 34.44/9.75 34.44/9.75 Q is empty. 34.44/9.75 34.44/9.75 ---------------------------------------- 34.44/9.75 34.44/9.75 (1) DependencyPairsProof (EQUIVALENT) 34.44/9.75 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 34.44/9.75 ---------------------------------------- 34.44/9.75 34.44/9.75 (2) 34.44/9.75 Obligation: 34.44/9.75 Q DP problem: 34.44/9.75 The TRS P consists of the following rules: 34.44/9.75 34.44/9.75 A(a(a(a(x1)))) -> B(b(x1)) 34.44/9.75 A(a(a(a(x1)))) -> B(x1) 34.44/9.75 B(b(a(a(x1)))) -> A(a(b(b(x1)))) 34.44/9.75 B(b(a(a(x1)))) -> A(b(b(x1))) 34.44/9.75 B(b(a(a(x1)))) -> B(b(x1)) 34.44/9.75 B(b(a(a(x1)))) -> B(x1) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> C(c(a(a(x1)))) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> C(a(a(x1))) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> A(a(x1)) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> A(x1) 34.44/9.75 B(b(b(b(x1)))) -> A(a(a(a(a(a(x1)))))) 34.44/9.75 B(b(b(b(x1)))) -> A(a(a(a(a(x1))))) 34.44/9.75 B(b(b(b(x1)))) -> A(a(a(a(x1)))) 34.44/9.75 B(b(b(b(x1)))) -> A(a(a(x1))) 34.44/9.75 B(b(b(b(x1)))) -> A(a(x1)) 34.44/9.75 B(b(b(b(x1)))) -> A(x1) 34.44/9.75 C(c(a(a(x1)))) -> B(b(a(a(c(c(x1)))))) 34.44/9.75 C(c(a(a(x1)))) -> B(a(a(c(c(x1))))) 34.44/9.75 C(c(a(a(x1)))) -> A(a(c(c(x1)))) 34.44/9.75 C(c(a(a(x1)))) -> A(c(c(x1))) 34.44/9.75 C(c(a(a(x1)))) -> C(c(x1)) 34.44/9.75 C(c(a(a(x1)))) -> C(x1) 34.44/9.75 34.44/9.75 The TRS R consists of the following rules: 34.44/9.75 34.44/9.75 a(a(a(a(x1)))) -> b(b(x1)) 34.44/9.75 b(b(a(a(x1)))) -> a(a(b(b(x1)))) 34.44/9.75 b(b(b(b(c(c(x1)))))) -> c(c(a(a(x1)))) 34.44/9.75 b(b(b(b(x1)))) -> a(a(a(a(a(a(x1)))))) 34.44/9.75 c(c(a(a(x1)))) -> b(b(a(a(c(c(x1)))))) 34.44/9.75 34.44/9.75 Q is empty. 34.44/9.75 We have to consider all minimal (P,Q,R)-chains. 34.44/9.75 ---------------------------------------- 34.44/9.75 34.44/9.75 (3) QDPOrderProof (EQUIVALENT) 34.44/9.75 We use the reduction pair processor [LPAR04,JAR06]. 34.44/9.75 34.44/9.75 34.44/9.75 The following pairs can be oriented strictly and are deleted. 34.44/9.75 34.44/9.75 B(b(b(b(c(c(x1)))))) -> C(a(a(x1))) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> A(a(x1)) 34.44/9.75 B(b(b(b(c(c(x1)))))) -> A(x1) 34.44/9.75 C(c(a(a(x1)))) -> C(x1) 34.44/9.75 The remaining pairs can at least be oriented weakly. 34.44/9.75 Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation: 34.44/9.75 34.44/9.75 POL( A_1(x_1) ) = x_1 + 1 34.44/9.75 POL( B_1(x_1) ) = x_1 + 1 popout

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all output return to SRS_Standard 2019-03-29 03.29