details

property	value
status	complete
benchmark	012.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n089.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.23072099686 seconds
cpu usage	5.866407931
max memory	4.8197632E8

stage attributes

key	value
output-size	7764
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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(and, true), true) -> true app(app(and, x), false) -> false app(app(and, false), y) -> false app(app(or, true), y) -> true app(app(or, x), true) -> true app(app(or, false), false) -> false app(app(forall, p), nil) -> true app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs)) app(app(forsome, p), nil) -> false app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), 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(and, true), true) -> true app(app(and, x), false) -> false app(app(and, false), y) -> false app(app(or, true), y) -> true app(app(or, x), true) -> true app(app(or, false), false) -> false app(app(forall, p), nil) -> true app(app(forall, p), app(app(cons, x), xs)) -> app(app(and, app(p, x)), app(app(forall, p), xs)) app(app(forsome, p), nil) -> false app(app(forsome, p), app(app(cons, x), xs)) -> app(app(or, app(p, x)), app(app(forsome, p), xs)) The set Q consists of the following terms: app(app(and, true), true) app(app(and, x0), false) app(app(and, false), x0) app(app(or, true), x0) app(app(or, x0), true) app(app(or, false), false) app(app(forall, x0), nil) app(app(forall, x0), app(app(cons, x1), x2)) app(app(forsome, x0), nil) app(app(forsome, 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(forall, p), app(app(cons, x), xs)) -> APP(app(and, app(p, x)), app(app(forall, p), xs)) APP(app(forall, p), app(app(cons, x), xs)) -> APP(and, app(p, x)) APP(app(forall, p), app(app(cons, x), xs)) -> APP(p, x) APP(app(forall, p), app(app(cons, x), xs)) -> APP(app(forall, p), xs) APP(app(forsome, p), app(app(cons, x), xs)) -> APP(app(or, app(p, x)), app(app(forsome, p), xs)) APP(app(forsome, p), app(app(cons, x), xs)) -> APP(or, app(p, x)) APP(app(forsome, p), app(app(cons, x), xs)) -> APP(p, x) popout

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job log

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actions

all output return to TRS Standard