details

property	value
status	complete
benchmark	hydra.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n190.star.cs.uiowa.edu
space	Applicative_first_order_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.21067 seconds
cpu usage	5.64684
user time	5.43271
system time	0.214125
max virtual memory	1.8345196E7
max residence set size	372500.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPApplicativeOrderProof [EQUIVALENT, 37 ms] (7) QDP (8) PisEmptyProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(f, app(app(cons, nil), y)) -> y app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z) app(app(app(copy, 0), y), z) -> app(f, z) app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z)) 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) Q is empty. ---------------------------------------- (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(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(app(copy, n), y), z) APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(app(copy, n), y) APP(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> APP(copy, n) APP(app(app(copy, 0), y), z) -> APP(f, z) APP(app(app(copy, app(s, x)), y), z) -> APP(app(app(copy, x), y), app(app(cons, app(f, y)), z)) APP(app(app(copy, app(s, x)), y), z) -> APP(app(copy, x), y) APP(app(app(copy, app(s, x)), y), z) -> APP(copy, x) APP(app(app(copy, app(s, x)), y), z) -> APP(app(cons, app(f, y)), z) APP(app(app(copy, app(s, x)), y), z) -> APP(cons, app(f, y)) APP(app(app(copy, app(s, x)), y), z) -> APP(f, y) 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(f, app(app(cons, nil), y)) -> y app(f, app(app(cons, app(f, app(app(cons, nil), y))), z)) -> app(app(app(copy, n), y), z) app(app(app(copy, 0), y), z) -> app(f, z) app(app(app(copy, app(s, x)), y), z) -> app(app(app(copy, x), y), app(app(cons, app(f, y)), z)) 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) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 18 less nodes. ---------------------------------------- popout

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job log

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actions

all output return to TRS Standard