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TRS Stand 20472 pair #381716960
details
property
value
status
complete
benchmark
ExConc_Zan97_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0455038547516 seconds
cpu usage
0.029555151
max memory
1433600.0
stage attributes
key
value
output-size
3059
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 f : [o] --> o g : [o] --> o h : [o] --> o n!6220!6220f : [o] --> o n!6220!6220h : [o] --> o f(X) => g(n!6220!6220h(n!6220!6220f(X))) h(X) => n!6220!6220h(X) f(X) => n!6220!6220f(X) activate(n!6220!6220h(X)) => h(activate(X)) activate(n!6220!6220f(X)) => f(activate(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(X) >? g(n!6220!6220h(n!6220!6220f(X))) h(X) >? n!6220!6220h(X) f(X) >? n!6220!6220f(X) activate(n!6220!6220h(X)) >? h(activate(X)) activate(n!6220!6220f(X)) >? f(activate(X)) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.2y0 f = \y0.3 + y0 g = \y0.y0 h = \y0.1 + y0 n!6220!6220f = \y0.2 + y0 n!6220!6220h = \y0.1 + y0 Using this interpretation, the requirements translate to: [[f(_x0)]] = 3 + x0 >= 3 + x0 = [[g(n!6220!6220h(n!6220!6220f(_x0)))]] [[h(_x0)]] = 1 + x0 >= 1 + x0 = [[n!6220!6220h(_x0)]] [[f(_x0)]] = 3 + x0 > 2 + x0 = [[n!6220!6220f(_x0)]] [[activate(n!6220!6220h(_x0))]] = 2 + 2x0 > 1 + 2x0 = [[h(activate(_x0))]] [[activate(n!6220!6220f(_x0))]] = 4 + 2x0 > 3 + 2x0 = [[f(activate(_x0))]] [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] We can thus remove the following rules: f(X) => n!6220!6220f(X) activate(n!6220!6220h(X)) => h(activate(X)) activate(n!6220!6220f(X)) => f(activate(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(X) >? g(n!6220!6220h(n!6220!6220f(X))) h(X) >? n!6220!6220h(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.3 + y0 f = \y0.3 + 3y0 g = \y0.y0 h = \y0.3 + 3y0 n!6220!6220f = \y0.y0 n!6220!6220h = \y0.y0 Using this interpretation, the requirements translate to: [[f(_x0)]] = 3 + 3x0 > x0 = [[g(n!6220!6220h(n!6220!6220f(_x0)))]] [[h(_x0)]] = 3 + 3x0 > x0 = [[n!6220!6220h(_x0)]] [[activate(_x0)]] = 3 + x0 > x0 = [[_x0]] We can thus remove the following rules: f(X) => g(n!6220!6220h(n!6220!6220f(X))) h(X) => n!6220!6220h(X) activate(X) => X All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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