details

property	value
status	complete
benchmark	#4.10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.7986240387 seconds
cpu usage	4.180413227
max memory	2.55377408E8

stage attributes

key	value
output-size	7930
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 63 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) UsableRulesProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(f, x), app(g, x)) -> app(app(f, 1), app(g, x)) app(g, 1) -> app(g, 0) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(f, x0), app(g, x0)) app(g, 1) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) app(app(app(app(filter2, true), x0), x1), x2) app(app(app(app(filter2, false), x0), x1), x2) ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(f, x), app(g, x)) -> APP(app(f, 1), app(g, x)) APP(app(f, x), app(g, x)) -> APP(f, 1) APP(g, 1) -> APP(g, 0) APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(cons, app(fun, x)), app(app(map, fun), xs)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(cons, app(fun, x)) APP(app(map, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(map, fun), app(app(cons, x), xs)) -> APP(app(map, fun), xs) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(fun, x)), fun), x), xs) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(app(filter2, app(fun, x)), fun), x) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(app(filter2, app(fun, x)), fun) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(filter2, app(fun, x)) APP(app(filter, fun), app(app(cons, x), xs)) -> APP(fun, x) APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(cons, x), app(app(filter, fun), xs)) APP(app(app(app(filter2, true), fun), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), fun), x), xs) -> APP(app(filter, fun), xs) APP(app(app(app(filter2, true), fun), x), xs) -> APP(filter, fun) APP(app(app(app(filter2, false), fun), x), xs) -> APP(app(filter, fun), xs) APP(app(app(app(filter2, false), fun), x), xs) -> APP(filter, fun) The TRS R consists of the following rules: app(app(f, x), app(g, x)) -> app(app(f, 1), app(g, x)) app(g, 1) -> app(g, 0) app(app(map, fun), nil) -> nil app(app(map, fun), app(app(cons, x), xs)) -> app(app(cons, app(fun, x)), app(app(map, fun), xs)) app(app(filter, fun), nil) -> nil app(app(filter, fun), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(fun, x)), fun), x), xs) app(app(app(app(filter2, true), fun), x), xs) -> app(app(cons, x), app(app(filter, fun), xs)) app(app(app(app(filter2, false), fun), x), xs) -> app(app(filter, fun), xs) The set Q consists of the following terms: app(app(f, x0), app(g, x0)) app(g, 1) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) popout

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

job log

popout

actions

all output return to TRS Inner 89993