details

property	value
status	complete
benchmark	10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n009.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	3.97670507431 seconds
cpu usage	12.022940879
max memory	7.34429184E8

stage attributes

key	value
output-size	6917
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) TransformationProof [EQUIVALENT, 0 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 38 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 212 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(y) -> y 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(y, 0))) C(c(c(y))) -> C(a(y, 0)) C(a(a(0, x), y)) -> C(c(c(0))) C(a(a(0, x), y)) -> C(c(0)) C(a(a(0, x), y)) -> C(0) The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(y) -> y 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 1 less node. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(c(y))) -> C(a(y, 0)) C(a(a(0, x), y)) -> C(c(c(0))) C(c(c(y))) -> C(c(a(y, 0))) C(a(a(0, x), y)) -> C(c(0)) The TRS R consists of the following rules: c(c(c(y))) -> c(c(a(y, 0))) c(a(a(0, x), y)) -> a(c(c(c(0))), y) c(y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) TransformationProof (EQUIVALENT) By narrowing [LPAR04] the rule C(a(a(0, x), y)) -> C(c(0)) at position [0] we obtained the following new rules [LPAR04]: (C(a(a(0, y0), y1)) -> C(0),C(a(a(0, y0), y1)) -> C(0)) popout

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

job log

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actions

all output return to TRS Standard