details

property	value
status	complete
benchmark	Ex24_GM04_Z.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n183.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	15.606 seconds
cpu usage	19.6073
user time	18.3124
system time	1.29489
max virtual memory	1.8413272E7
max residence set size	4089532.0

stage attributes

key	value
starexec-result	NO

output

NO proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 14 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) NonLoopProof [COMPLETE, 743 ms] (6) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(X, n__g(X), Y) -> f(activate(Y), activate(Y), activate(Y)) g(b) -> c b -> c g(X) -> n__g(X) activate(n__g(X)) -> g(X) activate(X) -> X 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(X, n__g(X), Y) -> F(activate(Y), activate(Y), activate(Y)) F(X, n__g(X), Y) -> ACTIVATE(Y) ACTIVATE(n__g(X)) -> G(X) The TRS R consists of the following rules: f(X, n__g(X), Y) -> f(activate(Y), activate(Y), activate(Y)) g(b) -> c b -> c g(X) -> n__g(X) activate(n__g(X)) -> g(X) activate(X) -> X 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 1 SCC with 2 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: F(X, n__g(X), Y) -> F(activate(Y), activate(Y), activate(Y)) The TRS R consists of the following rules: f(X, n__g(X), Y) -> f(activate(Y), activate(Y), activate(Y)) g(b) -> c b -> c g(X) -> n__g(X) activate(n__g(X)) -> g(X) activate(X) -> X Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) NonLoopProof (COMPLETE) By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP. We apply the theorem with m = 1, b = 0, σ' = [ ], and μ' = [x0 / c] on the rule F(c, n__g(c), n__g(b))[ ]^n[ ] -> F(c, n__g(c), n__g(b))[ ]^n[x0 / c] This rule is correct for the QDP as the following derivation shows: F(c, n__g(c), n__g(b))[ ]^n[ ] -> F(c, n__g(c), n__g(b))[ ]^n[x0 / c] by Equivalency by Simplifying Mu with mu1: [x0 / c] mu2: [ ] intermediate steps: Instantiate mu F(x0, n__g(x0), n__g(b))[ ]^n[ ] -> F(c, n__g(c), n__g(b))[ ]^n[ ] by Narrowing at position: [2] F(x0, n__g(x0), n__g(b))[ ]^n[ ] -> F(c, n__g(c), activate(n__g(b)))[ ]^n[ ] by Narrowing at position: [1] F(x0, n__g(x0), n__g(b))[ ]^n[ ] -> F(c, activate(n__g(c)), activate(n__g(b)))[ ]^n[ ] by Narrowing at position: [1,0,0] popout

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

job log

popout

actions

all output return to TRS Standard