details

property	value
status	complete
benchmark	innermost2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n068.star.cs.uiowa.edu
space	Mixed_innermost

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.19395709038 seconds
cpu usage	5.608687695
max memory	4.72420352E8

stage attributes

key	value
output-size	2452
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 47 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x)) f(a(x)) -> c a(b) -> d The set Q consists of the following terms: g(c, g(a(x0), x1)) f(a(x0)) a(b) ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b) = 0 POL(c) = 1 POL(d) = 0 POL(f(x_1)) = x_1 POL(g(x_1, x_2)) = 2*x_1 + 2*x_2 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a(b) -> d ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x)) f(a(x)) -> c The set Q consists of the following terms: g(c, g(a(x0), x1)) f(a(x0)) a(b) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: G(c, g(a(x), y)) -> G(f(a(b)), g(a(y), x)) G(c, g(a(x), y)) -> F(a(b)) G(c, g(a(x), y)) -> G(a(y), x) The TRS R consists of the following rules: g(c, g(a(x), y)) -> g(f(a(b)), g(a(y), x)) f(a(x)) -> c The set Q consists of the following terms: g(c, g(a(x0), x1)) f(a(x0)) popout

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

job log

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actions

all output return to TRS Innermost