details

property	value
status	complete
benchmark	Ex24_GM04_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	25.382 seconds
cpu usage	50.5164
user time	48.7108
system time	1.80561
max virtual memory	1.8611236E7
max residence set size	4788744.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) NonTerminationLoopProof [COMPLETE, 9844 ms] (7) NO (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a__f(X, g(X), Y) -> a__f(Y, Y, Y) a__g(b) -> c a__b -> c mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) mark(g(X)) -> a__g(mark(X)) mark(b) -> a__b mark(c) -> c a__f(X1, X2, X3) -> f(X1, X2, X3) a__g(X) -> g(X) a__b -> b Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) Obligation: Q DP problem: The TRS P consists of the following rules: A__F(X, g(X), Y) -> A__F(Y, Y, Y) MARK(f(X1, X2, X3)) -> A__F(X1, X2, X3) MARK(g(X)) -> A__G(mark(X)) MARK(g(X)) -> MARK(X) MARK(b) -> A__B The TRS R consists of the following rules: a__f(X, g(X), Y) -> a__f(Y, Y, Y) a__g(b) -> c a__b -> c mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) mark(g(X)) -> a__g(mark(X)) mark(b) -> a__b mark(c) -> c a__f(X1, X2, X3) -> f(X1, X2, X3) a__g(X) -> g(X) a__b -> b Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: A__F(X, g(X), Y) -> A__F(Y, Y, Y) The TRS R consists of the following rules: a__f(X, g(X), Y) -> a__f(Y, Y, Y) a__g(b) -> c a__b -> c mark(f(X1, X2, X3)) -> a__f(X1, X2, X3) mark(g(X)) -> a__g(mark(X)) mark(b) -> a__b popout

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job log

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actions

all output return to TRS Standard