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TRS Stand 20472 pair #381715277
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
n026.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
12.9450669289 seconds
cpu usage
25.041629815
max memory
4.557725696E9
stage attributes
key
value
output-size
4399
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 0 ms] (2) QDP (3) DependencyGraphProof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) QDP (5) NonLoopProof [COMPLETE, 683 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
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