details

property	value
status	complete
benchmark	2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	21.5281100273 seconds
cpu usage	51.99006508
max memory	1.690476544E9

stage attributes

key	value
output-size	6773
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 22 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 2830 ms] (12) QDP (13) DependencyGraphProof [EQUIVALENT, 0 ms] (14) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(c(b(x)))) -> a(1, b(c(x))) b(c(b(c(x)))) -> a(0, a(1, x)) a(0, x) -> c(c(x)) a(1, x) -> c(b(x)) 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(b(x)))) -> A(1, b(c(x))) C(c(c(b(x)))) -> B(c(x)) C(c(c(b(x)))) -> C(x) B(c(b(c(x)))) -> A(0, a(1, x)) B(c(b(c(x)))) -> A(1, x) A(0, x) -> C(c(x)) A(0, x) -> C(x) A(1, x) -> C(b(x)) A(1, x) -> B(x) The TRS R consists of the following rules: c(c(c(b(x)))) -> a(1, b(c(x))) b(c(b(c(x)))) -> a(0, a(1, x)) a(0, x) -> c(c(x)) a(1, x) -> c(b(x)) 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: C(c(c(b(x)))) -> B(c(x)) C(c(c(b(x)))) -> C(x) B(c(b(c(x)))) -> A(1, x) A(0, x) -> C(x) A(1, x) -> B(x) Used ordering: Polynomial interpretation [POLO]: POL(0) = 0 POL(1) = 0 POL(A(x_1, x_2)) = 1 + x_1 + x_2 POL(B(x_1)) = x_1 POL(C(x_1)) = x_1 POL(a(x_1, x_2)) = 2 + 2*x_1 + x_2 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 popout

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

job log

popout

actions

all output return to TRS Standard