details

property	value
status	complete
benchmark	8.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.77455 seconds
cpu usage	4.02542
user time	3.83972
system time	0.185704
max virtual memory	1.8475476E7
max residence set size	258432.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) MRRProof [EQUIVALENT, 18 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(0, y), x) -> y c(c(c(y))) -> c(c(a(a(c(b(0, y)), 0), 0))) a(y, 0) -> b(y, 0) 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: C(c(c(y))) -> C(c(a(a(c(b(0, y)), 0), 0))) C(c(c(y))) -> C(a(a(c(b(0, y)), 0), 0)) C(c(c(y))) -> A(a(c(b(0, y)), 0), 0) C(c(c(y))) -> A(c(b(0, y)), 0) C(c(c(y))) -> C(b(0, y)) C(c(c(y))) -> B(0, y) A(y, 0) -> B(y, 0) The TRS R consists of the following rules: b(b(0, y), x) -> y c(c(c(y))) -> c(c(a(a(c(b(0, y)), 0), 0))) a(y, 0) -> b(y, 0) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(c(y))) -> C(a(a(c(b(0, y)), 0), 0)) C(c(c(y))) -> C(c(a(a(c(b(0, y)), 0), 0))) The TRS R consists of the following rules: b(b(0, y), x) -> y c(c(c(y))) -> c(c(a(a(c(b(0, y)), 0), 0))) a(y, 0) -> b(y, 0) 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: C(c(c(y))) -> C(a(a(c(b(0, y)), 0), 0)) Used ordering: Polynomial interpretation [POLO]: popout

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

job log

popout

actions

all output return to TRS Standard