details

property	value
status	complete
benchmark	#3.49_rand.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	INVY_15

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.83899307251 seconds
cpu usage	4.102972576
max memory	2.73297408E8

stage attributes

key	value
output-size	6558
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given RelTRS could be proven: (0) RelTRS (1) RelTRStoQDPProof [SOUND, 0 ms] (2) QDP (3) MRRProof [EQUIVALENT, 56 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 41 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 0 ms] (11) QDP (12) PisEmptyProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) QDPOrderProof [EQUIVALENT, 0 ms] (16) QDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) YES ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: f(c(s(x), y)) -> f(c(x, s(y))) f(c(s(x), s(y))) -> g(c(x, y)) g(c(x, s(y))) -> g(c(s(x), y)) g(c(s(x), s(y))) -> f(c(x, y)) The relative TRS consists of the following S rules: rand(x) -> x rand(x) -> rand(s(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: f(c(s(x), y)) -> f(c(x, s(y))) f(c(s(x), s(y))) -> g(c(x, y)) g(c(x, s(y))) -> g(c(s(x), y)) g(c(s(x), s(y))) -> f(c(x, y)) The TRS R consists of the following rules: rand(x) -> x rand(x) -> rand(s(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 rules of the TRS R: rand(x) -> x Used ordering: Polynomial interpretation [POLO]: POL(c(x_1, x_2)) = x_1 + 2*x_2 POL(f(x_1)) = 2*x_1 POL(g(x_1)) = 2*x_1 POL(rand(x_1)) = 1 + 2*x_1 POL(s(x_1)) = x_1 ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: popout

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

job log

popout

actions

all output return to TRS Relat 75837