details

property	value
status	complete
benchmark	z081.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n184.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.16476 seconds
cpu usage	9.0875
user time	8.60652
system time	0.480982
max virtual memory	1.9608312E7
max residence set size	963892.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 48 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 1 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) MRRProof [EQUIVALENT, 27 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(c(a(x1))) -> a(b(x1)) b(b(b(x1))) -> c(a(c(x1))) c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) d(c(x1)) -> b(b(b(x1))) c(b(x1)) -> d(a(x1)) d(b(c(x1))) -> a(a(x1)) d(a(x1)) -> b(x1) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 POL(d(x_1)) = 2 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: b(c(a(x1))) -> a(b(x1)) b(b(b(x1))) -> c(a(c(x1))) d(b(c(x1))) -> a(a(x1)) d(a(x1)) -> b(x1) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) d(c(x1)) -> b(b(b(x1))) c(b(x1)) -> d(a(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: C(d(x1)) -> D(c(x1)) C(d(x1)) -> C(x1) C(d(b(x1))) -> D(c(c(x1))) C(d(b(x1))) -> C(c(x1)) C(d(b(x1))) -> C(x1) C(b(x1)) -> D(a(x1)) The TRS R consists of the following rules: c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) d(c(x1)) -> b(b(b(x1))) c(b(x1)) -> d(a(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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