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TRS Stand 20472 pair #381715715
details
property
value
status
complete
benchmark
Ex15_Luc06_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0619328022003 seconds
cpu usage
0.049792511
max memory
1835008.0
stage attributes
key
value
output-size
3654
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. a : [] --> o activate : [o] --> o f : [o] --> o g : [o] --> o n!6220!6220a : [] --> o n!6220!6220f : [o] --> o n!6220!6220g : [o] --> o f(n!6220!6220f(n!6220!6220a)) => f(n!6220!6220g(f(n!6220!6220a))) f(X) => n!6220!6220f(X) a => n!6220!6220a g(X) => n!6220!6220g(X) activate(n!6220!6220f(X)) => f(X) activate(n!6220!6220a) => a activate(n!6220!6220g(X)) => g(X) activate(X) => X We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(n!6220!6220f(n!6220!6220a)) >? f(n!6220!6220g(f(n!6220!6220a))) f(X) >? n!6220!6220f(X) a >? n!6220!6220a g(X) >? n!6220!6220g(X) activate(n!6220!6220f(X)) >? f(X) activate(n!6220!6220a) >? a activate(n!6220!6220g(X)) >? g(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 1 activate = \y0.3 + 3y0 f = \y0.y0 g = \y0.1 + y0 n!6220!6220a = 0 n!6220!6220f = \y0.y0 n!6220!6220g = \y0.y0 Using this interpretation, the requirements translate to: [[f(n!6220!6220f(n!6220!6220a))]] = 0 >= 0 = [[f(n!6220!6220g(f(n!6220!6220a)))]] [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] [[a]] = 1 > 0 = [[n!6220!6220a]] [[g(_x0)]] = 1 + x0 > x0 = [[n!6220!6220g(_x0)]] [[activate(n!6220!6220f(_x0))]] = 3 + 3x0 > x0 = [[f(_x0)]] [[activate(n!6220!6220a)]] = 3 > 1 = [[a]] [[activate(n!6220!6220g(_x0))]] = 3 + 3x0 > 1 + x0 = [[g(_x0)]] [[activate(_x0)]] = 3 + 3x0 > x0 = [[_x0]] We can thus remove the following rules: a => n!6220!6220a g(X) => n!6220!6220g(X) activate(n!6220!6220f(X)) => f(X) activate(n!6220!6220a) => a activate(n!6220!6220g(X)) => g(X) activate(X) => X We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] f#(n!6220!6220f(n!6220!6220a)) =#> f#(n!6220!6220g(f(n!6220!6220a))) 1] f#(n!6220!6220f(n!6220!6220a)) =#> f#(n!6220!6220a) Rules R_0: f(n!6220!6220f(n!6220!6220a)) => f(n!6220!6220g(f(n!6220!6220a))) f(X) => n!6220!6220f(X) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : This graph has no strongly connected components. By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
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