details

property	value
status	complete
benchmark	termMonTypes.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n178.star.cs.uiowa.edu
space	Applicative_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.48669 seconds
cpu usage	6.3359
user time	6.04312
system time	0.292787
max virtual memory	3.72417E7
max residence set size	394776.0

stage attributes

key	value
starexec-result	NO

output

NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) Overlay + Local Confluence [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof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, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 0 ms] (8) QDP (9) TransformationProof [EQUIVALENT, 0 ms] (10) QDP (11) DependencyGraphProof [EQUIVALENT, 0 ms] (12) AND (13) QDP (14) QDP (15) UsableRulesProof [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(f, x) -> x ap(ap(ap(g, x), y), ap(s, z)) -> ap(ap(ap(g, x), y), ap(ap(x, y), 0)) 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(f, x) -> x ap(ap(ap(g, x), y), ap(s, z)) -> ap(ap(ap(g, x), y), ap(ap(x, y), 0)) The set Q consists of the following terms: ap(f, x0) ap(ap(ap(g, x0), x1), ap(s, 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: AP(ap(ap(g, x), y), ap(s, z)) -> AP(ap(ap(g, x), y), ap(ap(x, y), 0)) AP(ap(ap(g, x), y), ap(s, z)) -> AP(ap(x, y), 0) AP(ap(ap(g, x), y), ap(s, z)) -> AP(x, y) The TRS R consists of the following rules: ap(f, x) -> x ap(ap(ap(g, x), y), ap(s, z)) -> ap(ap(ap(g, x), y), ap(ap(x, y), 0)) The set Q consists of the following terms: ap(f, x0) ap(ap(ap(g, x0), x1), ap(s, 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 1 less node. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: AP(ap(ap(g, x), y), ap(s, z)) -> AP(x, y) AP(ap(ap(g, x), y), ap(s, z)) -> AP(ap(ap(g, x), y), ap(ap(x, y), 0)) popout

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