details

property	value
status	complete
benchmark	rt1-3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n030.star.cs.uiowa.edu
space	Relative_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.61452794075 seconds
cpu usage	3.410964839
max memory	2.15904256E8

stage attributes

key	value
output-size	2509
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRStoQDPProof [SOUND, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 54 ms] (4) QDP (5) PisEmptyProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) The relative TRS consists of the following S rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(x)) ---------------------------------------- (1) RelTRStoQDPProof (SOUND) The relative termination problem is root-restricted. We can therefore treat it as a dependency pair problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) The TRS R consists of the following rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(x)) Q is empty. We have to consider all (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: t(f(x), g(y), f(z)) -> t(z, g(x), g(y)) t(g(x), g(y), f(z)) -> t(f(y), f(z), x) Used ordering: Polynomial interpretation [POLO]: POL(f(x_1)) = 2 + 2*x_1 POL(g(x_1)) = 2 + 2*x_1 POL(t(x_1, x_2, x_3)) = 2*x_1 + 2*x_2 + 2*x_3 ---------------------------------------- (4) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: f(g(x)) -> g(f(x)) g(f(x)) -> f(g(x)) f(f(x)) -> g(g(x)) g(g(x)) -> f(f(x)) Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (5) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. popout

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

job log

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actions

all output return to TRS Relat 75837