details

property	value
status	complete
benchmark	31.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	Applicative_first_order_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.17518 seconds
cpu usage	5.5149
user time	5.31202
system time	0.202881
max virtual memory	1.841096E7
max residence set size	364792.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 39 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) QDPOrderProof [EQUIVALENT, 0 ms] (7) QDP (8) PisEmptyProof [EQUIVALENT, 0 ms] (9) YES (10) QDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPSizeChangeProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z)) app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z)) app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), 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(app(:, app(app(:, x), y)), z) -> APP(app(:, x), app(app(:, y), z)) APP(app(:, app(app(:, x), y)), z) -> APP(app(:, y), z) APP(app(:, app(app(:, x), y)), z) -> APP(:, y) APP(app(:, app(app(+, x), y)), z) -> APP(app(+, app(app(:, x), z)), app(app(:, y), z)) APP(app(:, app(app(+, x), y)), z) -> APP(+, app(app(:, x), z)) APP(app(:, app(app(+, x), y)), z) -> APP(app(:, x), z) APP(app(:, app(app(+, x), y)), z) -> APP(:, x) APP(app(:, app(app(+, x), y)), z) -> APP(app(:, y), z) APP(app(:, app(app(+, x), y)), z) -> APP(:, y) APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(:, app(app(g, z), y)), app(app(+, x), a)) APP(app(:, z), app(app(+, x), app(f, y))) -> APP(:, app(app(g, z), y)) APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(g, z), y) APP(app(:, z), app(app(+, x), app(f, y))) -> APP(g, z) APP(app(:, z), app(app(+, x), app(f, y))) -> APP(app(+, x), a) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs)) APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x)) APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(app(filter2, app(f, x)), f), x), xs) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(filter2, app(f, x)), f), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter2, app(f, x)), f) APP(app(filter, f), app(app(cons, x), xs)) -> APP(filter2, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(cons, x), app(app(filter, f), xs)) APP(app(app(app(filter2, true), f), x), xs) -> APP(cons, x) APP(app(app(app(filter2, true), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, true), f), x), xs) -> APP(filter, f) APP(app(app(app(filter2, false), f), x), xs) -> APP(app(filter, f), xs) APP(app(app(app(filter2, false), f), x), xs) -> APP(filter, f) The TRS R consists of the following rules: app(app(:, app(app(:, x), y)), z) -> app(app(:, x), app(app(:, y), z)) app(app(:, app(app(+, x), y)), z) -> app(app(+, app(app(:, x), z)), app(app(:, y), z)) app(app(:, z), app(app(+, x), app(f, y))) -> app(app(:, app(app(g, z), y)), app(app(+, x), a)) app(app(map, f), nil) -> nil app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs)) app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(app(filter2, app(f, x)), f), x), xs) app(app(app(app(filter2, true), f), x), xs) -> app(app(cons, x), app(app(filter, f), xs)) app(app(app(app(filter2, false), f), x), xs) -> app(app(filter, f), xs) Q is empty. 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 Standard