details

property	value
status	complete
benchmark	04.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n127.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.09083008766 seconds
cpu usage	16.876615281
max memory	2.19205632E9

stage attributes

key	value
output-size	4922
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 1 ms] (2) QDP (3) MRRProof [EQUIVALENT, 54 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 15 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 25 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(x1))) -> b(a(x1)) b(a(a(x1))) -> a(a(a(b(x1)))) a(c(x1)) -> c(b(x1)) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(x1))) -> B(a(x1)) A(a(b(x1))) -> A(x1) B(a(a(x1))) -> A(a(a(b(x1)))) B(a(a(x1))) -> A(a(b(x1))) B(a(a(x1))) -> A(b(x1)) B(a(a(x1))) -> B(x1) A(c(x1)) -> B(x1) The TRS R consists of the following rules: a(a(b(x1))) -> b(a(x1)) b(a(a(x1))) -> a(a(a(b(x1)))) a(c(x1)) -> c(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) MRRProof (EQUIVALENT) 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. Strictly oriented dependency pairs: A(c(x1)) -> B(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 2*x_1 POL(B(x_1)) = 2*x_1 POL(a(x_1)) = x_1 POL(b(x_1)) = x_1 POL(c(x_1)) = 2 + x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(a(b(x1))) -> B(a(x1)) A(a(b(x1))) -> A(x1) B(a(a(x1))) -> A(a(a(b(x1)))) B(a(a(x1))) -> A(a(b(x1))) B(a(a(x1))) -> A(b(x1)) B(a(a(x1))) -> B(x1) The TRS R consists of the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to SRS Standard