details

property	value
status	complete
benchmark	z089.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.68232 seconds
cpu usage	11.4912
user time	10.9848
system time	0.506387
max virtual memory	5.9402812E7
max residence set size	1320484.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) MRRProof [EQUIVALENT, 52 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 139 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(a(x1)))) -> a(c(a(b(x1)))) a(c(x1)) -> c(c(a(x1))) c(c(c(x1))) -> b(c(b(x1))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(a(x1)))) -> b(a(c(a(x1)))) c(a(x1)) -> a(c(c(x1))) c(c(c(x1))) -> b(c(b(x1))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(b(a(x1)))) -> A(c(a(x1))) A(b(b(a(x1)))) -> C(a(x1)) C(a(x1)) -> A(c(c(x1))) C(a(x1)) -> C(c(x1)) C(a(x1)) -> C(x1) C(c(c(x1))) -> C(b(x1)) The TRS R consists of the following rules: a(b(b(a(x1)))) -> b(a(c(a(x1)))) c(a(x1)) -> a(c(c(x1))) c(c(c(x1))) -> b(c(b(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(b(a(x1)))) -> C(a(x1)) C(a(x1)) -> A(c(c(x1))) A(b(b(a(x1)))) -> A(c(a(x1))) C(a(x1)) -> C(c(x1)) C(a(x1)) -> C(x1) The TRS R consists of the following rules: a(b(b(a(x1)))) -> b(a(c(a(x1)))) popout

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