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	n137.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.99122 seconds
cpu usage	4.49823
user time	4.34131
system time	0.156916
max virtual memory	1.84176E7
max residence set size	259728.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 19 ms] (2) QDP (3) DependencyGraphProof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) 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) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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job log

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actions

all output return to TRS Innermost