details

property	value
status	complete
benchmark	secr8.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n061.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.94817399979 seconds
cpu usage	20.180518324
max memory	1.352609792E9

stage attributes

key	value
output-size	6293
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 13 ms] (2) QDP (3) MRRProof [EQUIVALENT, 62 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 90 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 93 ms] (10) QDP (11) PisEmptyProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(b(x1))) -> a(x1) a(a(x1)) -> a(b(a(x1))) b(c(x1)) -> c(a(a(x1))) a(c(x1)) -> c(b(b(x1))) a(a(a(x1))) -> b(a(a(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: B(b(b(x1))) -> A(x1) A(a(x1)) -> A(b(a(x1))) A(a(x1)) -> B(a(x1)) B(c(x1)) -> A(a(x1)) B(c(x1)) -> A(x1) A(c(x1)) -> B(b(x1)) A(c(x1)) -> B(x1) A(a(a(x1))) -> B(a(a(x1))) The TRS R consists of the following rules: b(b(b(x1))) -> a(x1) a(a(x1)) -> a(b(a(x1))) b(c(x1)) -> c(a(a(x1))) a(c(x1)) -> c(b(b(x1))) a(a(a(x1))) -> b(a(a(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: B(c(x1)) -> A(a(x1)) B(c(x1)) -> A(x1) A(c(x1)) -> B(b(x1)) A(c(x1)) -> B(x1) Used ordering: Polynomial interpretation [POLO]: POL(A(x_1)) = x_1 POL(B(x_1)) = 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: popout

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

job log

popout

actions

all output return to SRS Standard