details

property	value
status	complete
benchmark	test9.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n180.star.cs.uiowa.edu
space	Strategy_removed_mixed_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.99982 seconds
cpu usage	4.4745
user time	4.26776
system time	0.206744
max virtual memory	1.8476724E7
max residence set size	316168.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof [COMPLETE, 7 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, X, X) g(X, Y) -> X g(X, Y) -> Y 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: F(0, 1, X) -> H(X, X) H(0, X) -> F(0, X, X) The TRS R consists of the following rules: f(0, 1, X) -> h(X, X) h(0, X) -> f(0, X, X) g(X, Y) -> X g(X, Y) -> Y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = H(g(X, 0), g(1, Y)) evaluates to t =H(g(1, Y), g(1, Y)) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [X / 1, Y / 0] -------------------------------------------------------------------------------- Rewriting sequence H(g(1, 0), g(1, 0)) -> H(0, g(1, 0)) with rule g(X, Y') -> Y' at position [0] and matcher [X / 1, Y' / 0] H(0, g(1, 0)) -> F(0, g(1, 0), g(1, 0)) with rule H(0, X') -> F(0, X', X') at position [] and matcher [X' / g(1, 0)] F(0, g(1, 0), g(1, 0)) -> F(0, 1, g(1, 0)) with rule g(X', Y) -> X' at position [1] and matcher [X' / 1, Y / 0] F(0, 1, g(1, 0)) -> H(g(1, 0), g(1, 0)) with rule F(0, 1, X) -> H(X, X) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (4) NO popout

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all output return to TRS Standard