details

property	value
status	complete
benchmark	2.61.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n083.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.11437106133 seconds
cpu usage	5.538318969
max memory	4.6913536E8

stage attributes

key	value
output-size	2041
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 33 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(j(x, y), y) -> g(f(x, k(y))) f(x, h1(y, z)) -> h2(0, x, h1(y, z)) g(h2(x, y, h1(z, u))) -> h2(s(x), y, h1(z, u)) h2(x, j(y, h1(z, u)), h1(z, u)) -> h2(s(x), y, h1(s(z), u)) i(f(x, h(y))) -> y i(h2(s(x), y, h1(x, z))) -> z k(h(x)) -> h1(0, x) k(h1(x, y)) -> h1(s(x), y) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Knuth-Bendix order [KBO] with precedence:h_1 > i_1 > s_1 > 0 > f_2 > g_1 > j_2 > h2_3 > k_1 > h1_2 and weight map: 0=5 g_1=4 k_1=4 s_1=4 i_1=1 h_1=1 j_2=7 f_2=5 h2_3=0 h1_2=0 The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(j(x, y), y) -> g(f(x, k(y))) f(x, h1(y, z)) -> h2(0, x, h1(y, z)) g(h2(x, y, h1(z, u))) -> h2(s(x), y, h1(z, u)) h2(x, j(y, h1(z, u)), h1(z, u)) -> h2(s(x), y, h1(s(z), u)) i(f(x, h(y))) -> y i(h2(s(x), y, h1(x, z))) -> z k(h(x)) -> h1(0, x) k(h1(x, y)) -> h1(s(x), y) ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES popout

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all output return to TRS Standard