details

property	value
status	complete
benchmark	#4.8.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n184.star.cs.uiowa.edu
space	Applicative_AG01_innermost

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.14562106133 seconds
cpu usage	8.144576698
max memory	4.41860096E8

stage attributes

key	value
output-size	17763
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 20 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) ATransformationProof [EQUIVALENT, 0 ms] (9) QDP (10) QReductionProof [EQUIVALENT, 0 ms] (11) QDP (12) QDPSizeChangeProof [EQUIVALENT, 0 ms] (13) YES (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) ATransformationProof [EQUIVALENT, 0 ms] (18) QDP (19) QReductionProof [EQUIVALENT, 0 ms] (20) QDP (21) TransformationProof [EQUIVALENT, 0 ms] (22) QDP (23) DependencyGraphProof [EQUIVALENT, 0 ms] (24) TRUE (25) QDP (26) QDPSizeChangeProof [EQUIVALENT, 0 ms] (27) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(plus, x), 0) -> x app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y)) app(app(app(f, 0), app(s, 0)), x) -> app(app(app(f, x), app(app(plus, x), x)), x) 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(plus, x0), 0) app(app(plus, x0), app(s, x1)) app(app(app(f, 0), app(s, 0)), x0) 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(plus, x), app(s, y)) -> APP(s, app(app(plus, x), y)) APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y) APP(app(app(f, 0), app(s, 0)), x) -> APP(app(app(f, x), app(app(plus, x), x)), x) APP(app(app(f, 0), app(s, 0)), x) -> APP(app(f, x), app(app(plus, x), x)) APP(app(app(f, 0), app(s, 0)), x) -> APP(f, x) APP(app(app(f, 0), app(s, 0)), x) -> APP(app(plus, x), x) APP(app(app(f, 0), app(s, 0)), x) -> APP(plus, x) 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) popout

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

job log

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actions

all output return to TRS Innermost