details

property	value
status	complete
benchmark	028.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	AProVE23
configuration	standard
runtime (wallclock)	1.7653400898 seconds
cpu usage	4.43392884
max memory	2.62742016E8

stage attributes

key	value
output-size	5561
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPSizeChangeProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) Q is empty. ---------------------------------------- (1) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) The set Q consists of the following terms: app(app(app(consif, true), x0), x1) app(app(app(consif, false), x0), x1) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, x1), x2)) ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: APP(app(app(consif, true), x), ys) -> APP(app(cons, x), ys) APP(app(app(consif, true), x), ys) -> APP(cons, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(consif, app(f, x)), x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(consif, app(f, x)) APP(app(filter, f), app(app(cons, x), xs)) -> APP(f, x) APP(app(filter, f), app(app(cons, x), xs)) -> APP(app(filter, f), xs) The TRS R consists of the following rules: app(app(app(consif, true), x), ys) -> app(app(cons, x), ys) app(app(app(consif, false), x), ys) -> ys app(app(filter, f), nil) -> nil app(app(filter, f), app(app(cons, x), xs)) -> app(app(app(consif, app(f, x)), x), app(app(filter, f), xs)) The set Q consists of the following terms: app(app(app(consif, true), x0), x1) app(app(app(consif, false), x0), x1) app(app(filter, x0), nil) app(app(filter, x0), app(app(cons, 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 Standard