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TRS Stand 20472 pair #381715905
details
property
value
status
complete
benchmark
Ex25_Luc06_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.128557920456 seconds
cpu usage
0.126158627
max memory
3751936.0
stage attributes
key
value
output-size
8764
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. activate : [o] --> o c : [o] --> o d : [o] --> o f : [o] --> o g : [o] --> o h : [o] --> o n!6220!6220d : [o] --> o n!6220!6220f : [o] --> o n!6220!6220g : [o] --> o f(f(X)) => c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) c(X) => d(activate(X)) h(X) => c(n!6220!6220d(X)) f(X) => n!6220!6220f(X) g(X) => n!6220!6220g(X) d(X) => n!6220!6220d(X) activate(n!6220!6220f(X)) => f(activate(X)) activate(n!6220!6220g(X)) => g(X) activate(n!6220!6220d(X)) => d(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(f(X)) >? c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) c(X) >? d(activate(X)) h(X) >? c(n!6220!6220d(X)) f(X) >? n!6220!6220f(X) g(X) >? n!6220!6220g(X) d(X) >? n!6220!6220d(X) activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220g(X)) >? g(X) activate(n!6220!6220d(X)) >? d(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.y0 c = \y0.y0 d = \y0.y0 f = \y0.y0 g = \y0.y0 h = \y0.3 + 3y0 n!6220!6220d = \y0.y0 n!6220!6220f = \y0.y0 n!6220!6220g = \y0.y0 Using this interpretation, the requirements translate to: [[f(f(_x0))]] = x0 >= x0 = [[c(n!6220!6220f(n!6220!6220g(n!6220!6220f(_x0))))]] [[c(_x0)]] = x0 >= x0 = [[d(activate(_x0))]] [[h(_x0)]] = 3 + 3x0 > x0 = [[c(n!6220!6220d(_x0))]] [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] [[g(_x0)]] = x0 >= x0 = [[n!6220!6220g(_x0)]] [[d(_x0)]] = x0 >= x0 = [[n!6220!6220d(_x0)]] [[activate(n!6220!6220f(_x0))]] = x0 >= x0 = [[f(activate(_x0))]] [[activate(n!6220!6220g(_x0))]] = x0 >= x0 = [[g(_x0)]] [[activate(n!6220!6220d(_x0))]] = x0 >= x0 = [[d(_x0)]] [[activate(_x0)]] = x0 >= x0 = [[_x0]] We can thus remove the following rules: h(X) => c(n!6220!6220d(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#(f(X)) =#> c#(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) 1] c#(X) =#> d#(activate(X)) 2] c#(X) =#> activate#(X) 3] activate#(n!6220!6220f(X)) =#> f#(activate(X)) 4] activate#(n!6220!6220f(X)) =#> activate#(X) 5] activate#(n!6220!6220g(X)) =#> g#(X) 6] activate#(n!6220!6220d(X)) =#> d#(X) Rules R_0: f(f(X)) => c(n!6220!6220f(n!6220!6220g(n!6220!6220f(X)))) c(X) => d(activate(X)) f(X) => n!6220!6220f(X) g(X) => n!6220!6220g(X) d(X) => n!6220!6220d(X) activate(n!6220!6220f(X)) => f(activate(X))
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