details

property	value
status	complete
benchmark	#4.3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n009.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.36736392975 seconds
cpu usage	6.236475076
max memory	4.82066432E8

stage attributes

key	value
output-size	8930
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 65 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0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(app(f, app(app(g, x), y)), x), z) -> app(app(app(f, z), z), z) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(g, x0), x1) 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(app(f, app(app(g, x), y)), x), z) -> APP(app(app(f, z), z), z) APP(app(app(f, app(app(g, x), y)), x), z) -> APP(app(f, z), z) APP(app(app(f, app(app(g, x), y)), x), z) -> APP(f, z) 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(app(f, app(app(g, x), y)), x), z) -> app(app(app(f, z), z), z) app(app(g, x), y) -> x app(app(g, x), y) -> y 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(g, x0), x1) app(app(map, x0), nil) app(app(map, x0), app(app(cons, x1), x2)) app(app(filter, x0), nil) popout

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job log

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actions

all output return to TRS Innermost