details

property	value
status	complete
benchmark	2.52.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n023.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.72140812874 seconds
cpu usage	4.0138255
max memory	2.36097536E8

stage attributes

key	value
output-size	2104
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 157 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4)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(x, 0, 0) -> s(x) f(0, y, 0) -> s(y) f(0, 0, z) -> s(z) f(s(0), y, z) -> f(0, s(y), s(z)) f(s(x), s(y), 0) -> f(x, y, s(0)) f(s(x), 0, s(z)) -> f(x, s(0), z) f(0, s(0), s(0)) -> s(s(0)) f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z)) f(0, s(s(y)), s(0)) -> f(0, y, s(0)) f(0, s(0), s(s(z))) -> f(0, s(0), z) f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: f_3 > [0, s_1] Status: f_3: [1,2,3] 0: multiset status s_1: [1] With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(x, 0, 0) -> s(x) f(0, y, 0) -> s(y) f(0, 0, z) -> s(z) f(s(0), y, z) -> f(0, s(y), s(z)) f(s(x), s(y), 0) -> f(x, y, s(0)) f(s(x), 0, s(z)) -> f(x, s(0), z) f(0, s(0), s(0)) -> s(s(0)) f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z)) f(0, s(s(y)), s(0)) -> f(0, y, s(0)) f(0, s(0), s(s(z))) -> f(0, s(0), z) f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z))) ---------------------------------------- (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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