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TRS Stand 20472 pair #381716735
details
property
value
status
complete
benchmark
Ex25_Luc06_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.093722820282 seconds
cpu usage
0.09018252
max memory
2600960.0
stage attributes
key
value
output-size
7095
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. a!6220!6220c : [o] --> o a!6220!6220f : [o] --> o a!6220!6220h : [o] --> o c : [o] --> o d : [o] --> o f : [o] --> o g : [o] --> o h : [o] --> o mark : [o] --> o a!6220!6220f(f(X)) => a!6220!6220c(f(g(f(X)))) a!6220!6220c(X) => d(X) a!6220!6220h(X) => a!6220!6220c(d(X)) mark(f(X)) => a!6220!6220f(mark(X)) mark(c(X)) => a!6220!6220c(X) mark(h(X)) => a!6220!6220h(mark(X)) mark(g(X)) => g(X) mark(d(X)) => d(X) a!6220!6220f(X) => f(X) a!6220!6220c(X) => c(X) a!6220!6220h(X) => h(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]): a!6220!6220f(f(X)) >? a!6220!6220c(f(g(f(X)))) a!6220!6220c(X) >? d(X) a!6220!6220h(X) >? a!6220!6220c(d(X)) mark(f(X)) >? a!6220!6220f(mark(X)) mark(c(X)) >? a!6220!6220c(X) mark(h(X)) >? a!6220!6220h(mark(X)) mark(g(X)) >? g(X) mark(d(X)) >? d(X) a!6220!6220f(X) >? f(X) a!6220!6220c(X) >? c(X) a!6220!6220h(X) >? h(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220c = \y0.y0 a!6220!6220f = \y0.2y0 a!6220!6220h = \y0.2 + y0 c = \y0.y0 d = \y0.y0 f = \y0.2y0 g = \y0.y0 h = \y0.1 + y0 mark = \y0.2y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(f(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220c(f(g(f(_x0))))]] [[a!6220!6220c(_x0)]] = x0 >= x0 = [[d(_x0)]] [[a!6220!6220h(_x0)]] = 2 + x0 > x0 = [[a!6220!6220c(d(_x0))]] [[mark(f(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220f(mark(_x0))]] [[mark(c(_x0))]] = 2x0 >= x0 = [[a!6220!6220c(_x0)]] [[mark(h(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220h(mark(_x0))]] [[mark(g(_x0))]] = 2x0 >= x0 = [[g(_x0)]] [[mark(d(_x0))]] = 2x0 >= x0 = [[d(_x0)]] [[a!6220!6220f(_x0)]] = 2x0 >= 2x0 = [[f(_x0)]] [[a!6220!6220c(_x0)]] = x0 >= x0 = [[c(_x0)]] [[a!6220!6220h(_x0)]] = 2 + x0 > 1 + x0 = [[h(_x0)]] We can thus remove the following rules: a!6220!6220h(X) => a!6220!6220c(d(X)) a!6220!6220h(X) => h(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]): a!6220!6220f(f(X)) >? a!6220!6220c(f(g(f(X)))) a!6220!6220c(X) >? d(X) mark(f(X)) >? a!6220!6220f(mark(X)) mark(c(X)) >? a!6220!6220c(X) mark(h(X)) >? a!6220!6220h(mark(X)) mark(g(X)) >? g(X) mark(d(X)) >? d(X) a!6220!6220f(X) >? f(X) a!6220!6220c(X) >? c(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements:
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