details

property	value
status	complete
benchmark	003.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.62425494194 seconds
cpu usage	7.394915243
max memory	5.16841472E8

stage attributes

key	value
output-size	6401
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 24 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 0 ms] (8) QDP (9) MNOCProof [EQUIVALENT, 0 ms] (10) QDP (11) NonTerminationLoopProof [COMPLETE, 0 ms] (12) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(app(app(until, p), f), x) -> app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x))) 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(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(app(app(until, p), f), x) -> app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x))) The set Q consists of the following terms: app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(app(app(until, x0), 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(until, p), f), x) -> APP(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x))) APP(app(app(until, p), f), x) -> APP(app(if, app(p, x)), x) APP(app(app(until, p), f), x) -> APP(if, app(p, x)) APP(app(app(until, p), f), x) -> APP(p, x) APP(app(app(until, p), f), x) -> APP(app(app(until, p), f), app(f, x)) APP(app(app(until, p), f), x) -> APP(f, x) The TRS R consists of the following rules: app(app(app(if, true), x), y) -> x app(app(app(if, false), x), y) -> y app(app(app(until, p), f), x) -> app(app(app(if, app(p, x)), x), app(app(app(until, p), f), app(f, x))) The set Q consists of the following terms: app(app(app(if, true), x0), x1) app(app(app(if, false), x0), x1) app(app(app(until, x0), x1), x2) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- popout

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all output return to TRS Standard