details

property	value
status	complete
benchmark	nonTermF.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n079.star.cs.uiowa.edu
space	Applicative_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	6.58402204514 seconds
cpu usage	20.10633733
max memory	1.143054336E9

stage attributes

key	value
output-size	19730
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 ms] (6) QDP (7) TransformationProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) TransformationProof [EQUIVALENT, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 20 ms] (14) QDP (15) MNOCProof [EQUIVALENT, 0 ms] (16) QDP (17) NonTerminationLoopProof [COMPLETE, 0 ms] (18) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil)) ap(ap(ap(foldr, g), h), nil) -> h ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), 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: ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil)) ap(ap(ap(foldr, g), h), nil) -> h ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), xs)) The set Q consists of the following terms: ap(ap(f, x0), x0) ap(ap(ap(foldr, x0), x1), nil) ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3)) ---------------------------------------- (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: AP(ap(f, x), x) -> AP(ap(x, ap(f, x)), ap(ap(cons, x), nil)) AP(ap(f, x), x) -> AP(x, ap(f, x)) AP(ap(f, x), x) -> AP(ap(cons, x), nil) AP(ap(f, x), x) -> AP(cons, x) AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(g, x), ap(ap(ap(foldr, g), h), xs)) AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(g, x) AP(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> AP(ap(ap(foldr, g), h), xs) The TRS R consists of the following rules: ap(ap(f, x), x) -> ap(ap(x, ap(f, x)), ap(ap(cons, x), nil)) ap(ap(ap(foldr, g), h), nil) -> h ap(ap(ap(foldr, g), h), ap(ap(cons, x), xs)) -> ap(ap(g, x), ap(ap(ap(foldr, g), h), xs)) The set Q consists of the following terms: ap(ap(f, x0), x0) ap(ap(ap(foldr, x0), x1), nil) ap(ap(ap(foldr, x0), x1), ap(ap(cons, x2), x3)) popout

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actions

all output return to TRS Standard