details

property	value
status	complete
benchmark	z111.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n173.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.08683 seconds
cpu usage	16.0037
user time	15.2474
system time	0.756352
max virtual memory	1.9945648E7
max residence set size	2175540.0

stage attributes

key	value
starexec-result	YES

output

15.41/4.99 YES 15.85/5.01 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 15.85/5.01 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 15.85/5.01 15.85/5.01 15.85/5.01 Termination w.r.t. Q of the given QTRS could be proven: 15.85/5.01 15.85/5.01 (0) QTRS 15.85/5.01 (1) QTRSRRRProof [EQUIVALENT, 69 ms] 15.85/5.01 (2) QTRS 15.85/5.01 (3) DependencyPairsProof [EQUIVALENT, 9 ms] 15.85/5.01 (4) QDP 15.85/5.01 (5) DependencyGraphProof [EQUIVALENT, 0 ms] 15.85/5.01 (6) QDP 15.85/5.01 (7) MRRProof [EQUIVALENT, 39 ms] 15.85/5.01 (8) QDP 15.85/5.01 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 15.85/5.01 (10) TRUE 15.85/5.01 15.85/5.01 15.85/5.01 ---------------------------------------- 15.85/5.01 15.85/5.01 (0) 15.85/5.01 Obligation: 15.85/5.01 Q restricted rewrite system: 15.85/5.01 The TRS R consists of the following rules: 15.85/5.01 15.85/5.01 a(a(x1)) -> b(c(c(c(x1)))) 15.85/5.01 b(c(x1)) -> d(d(d(d(x1)))) 15.85/5.01 a(x1) -> d(c(d(x1))) 15.85/5.01 b(b(x1)) -> c(c(c(x1))) 15.85/5.01 c(c(x1)) -> d(d(d(x1))) 15.85/5.01 c(d(d(x1))) -> a(x1) 15.85/5.01 15.85/5.01 Q is empty. 15.85/5.01 15.85/5.01 ---------------------------------------- 15.85/5.01 15.85/5.01 (1) QTRSRRRProof (EQUIVALENT) 15.85/5.01 Used ordering: 15.85/5.01 Polynomial interpretation [POLO]: 15.85/5.01 15.85/5.01 POL(a(x_1)) = 25 + x_1 15.85/5.01 POL(b(x_1)) = 17 + x_1 15.85/5.01 POL(c(x_1)) = 11 + x_1 15.85/5.01 POL(d(x_1)) = 7 + x_1 15.85/5.01 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 15.85/5.03 15.85/5.03 b(b(x1)) -> c(c(c(x1))) 15.85/5.03 c(c(x1)) -> d(d(d(x1))) 15.85/5.03 15.85/5.03 15.85/5.03 15.85/5.03 15.85/5.03 ---------------------------------------- 15.85/5.03 15.85/5.03 (2) 15.85/5.03 Obligation: 15.85/5.03 Q restricted rewrite system: 15.85/5.03 The TRS R consists of the following rules: 15.85/5.03 15.85/5.03 a(a(x1)) -> b(c(c(c(x1)))) 15.85/5.03 b(c(x1)) -> d(d(d(d(x1)))) 15.85/5.03 a(x1) -> d(c(d(x1))) 15.85/5.03 c(d(d(x1))) -> a(x1) 15.85/5.03 15.85/5.03 Q is empty. 15.85/5.03 15.85/5.03 ---------------------------------------- 15.85/5.03 15.85/5.03 (3) DependencyPairsProof (EQUIVALENT) 15.85/5.03 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 15.85/5.03 ---------------------------------------- 15.85/5.03 15.85/5.03 (4) 15.85/5.03 Obligation: 15.85/5.03 Q DP problem: 15.85/5.03 The TRS P consists of the following rules: 15.85/5.03 15.85/5.03 A(a(x1)) -> B(c(c(c(x1)))) 15.85/5.03 A(a(x1)) -> C(c(c(x1))) 15.85/5.03 A(a(x1)) -> C(c(x1)) 15.85/5.03 A(a(x1)) -> C(x1) 15.85/5.03 A(x1) -> C(d(x1)) 15.85/5.03 C(d(d(x1))) -> A(x1) 15.85/5.03 15.85/5.03 The TRS R consists of the following rules: 15.85/5.03 15.85/5.03 a(a(x1)) -> b(c(c(c(x1)))) 15.85/5.03 b(c(x1)) -> d(d(d(d(x1)))) 15.85/5.03 a(x1) -> d(c(d(x1))) 15.85/5.03 c(d(d(x1))) -> a(x1) 15.85/5.03 15.85/5.03 Q is empty. 15.85/5.03 We have to consider all minimal (P,Q,R)-chains. 15.85/5.03 ---------------------------------------- 15.85/5.03 15.85/5.03 (5) DependencyGraphProof (EQUIVALENT) 15.85/5.03 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 15.85/5.03 ---------------------------------------- popout

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