details

property	value
status	complete
benchmark	#4.3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.85231 seconds
cpu usage	3.98956
user time	3.79486
system time	0.194694
max virtual memory	1.841096E7
max residence set size	301796.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) NonTerminationLoopProof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, 0 ms] (4) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(g(x, y), x, z) -> f(z, z, z) 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(g(x, y), x, z) -> F(z, z, z) The TRS R consists of the following rules: f(g(x, y), x, z) -> f(z, z, z) 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 = F(g(x', y), g(x', y'), z) evaluates to t =F(z, z, z) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [y' / y, z / g(x', y)] -------------------------------------------------------------------------------- Rewriting sequence F(g(x', y), g(x', y), g(x', y)) -> F(g(x', y), x', g(x', y)) with rule g(x'', y') -> x'' at position [1] and matcher [x'' / x', y' / y] F(g(x', y), x', g(x', y)) -> F(g(x', y), g(x', y), g(x', y)) with rule F(g(x, y), x, z) -> F(z, z, z) 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

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard