details

property	value
status	complete
benchmark	z086.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n002.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	4.99053096771 seconds
cpu usage	16.543507009
max memory	1.2389376E9

stage attributes

key	value
output-size	4311
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 11 ms] (4) QDP (5) MRRProof [EQUIVALENT, 55 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 50 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(x1)) -> c(b(x1)) b(b(x1)) -> c(a(x1)) c(c(x1)) -> b(a(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(a(x1)) -> b(c(x1)) b(b(x1)) -> a(c(x1)) c(c(x1)) -> a(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(a(x1)) -> B(c(x1)) A(a(x1)) -> C(x1) B(b(x1)) -> A(c(x1)) B(b(x1)) -> C(x1) C(c(x1)) -> A(b(x1)) C(c(x1)) -> B(x1) The TRS R consists of the following rules: a(a(x1)) -> b(c(x1)) b(b(x1)) -> a(c(x1)) c(c(x1)) -> a(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) 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(a(x1)) -> C(x1) B(b(x1)) -> C(x1) C(c(x1)) -> A(b(x1)) C(c(x1)) -> B(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = 2*x_1 POL(B(x_1)) = 2*x_1 POL(C(x_1)) = 3 + 2*x_1 popout

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

job log

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all output return to SRS Standard