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TRS Stand 20472 pair #381716145
details
property
value
status
complete
benchmark
Ex23_Luc06_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n097.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0660548210144 seconds
cpu usage
0.043010176
max memory
1736704.0
stage attributes
key
value
output-size
3327
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 a!6220!6220f : [o] --> o c : [o] --> o f : [o] --> o g : [o] --> o mark : [o] --> o a!6220!6220f(f(a)) => c(f(g(f(a)))) mark(f(X)) => a!6220!6220f(mark(X)) mark(a) => a mark(c(X)) => c(X) mark(g(X)) => g(mark(X)) a!6220!6220f(X) => f(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(a)) >? c(f(g(f(a)))) mark(f(X)) >? a!6220!6220f(mark(X)) mark(a) >? a mark(c(X)) >? c(X) mark(g(X)) >? g(mark(X)) a!6220!6220f(X) >? f(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 0 a!6220!6220f = \y0.1 + y0 c = \y0.y0 f = \y0.1 + y0 g = \y0.y0 mark = \y0.2 + 2y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(f(a))]] = 2 >= 2 = [[c(f(g(f(a))))]] [[mark(f(_x0))]] = 4 + 2x0 > 3 + 2x0 = [[a!6220!6220f(mark(_x0))]] [[mark(a)]] = 2 > 0 = [[a]] [[mark(c(_x0))]] = 2 + 2x0 > x0 = [[c(_x0)]] [[mark(g(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[g(mark(_x0))]] [[a!6220!6220f(_x0)]] = 1 + x0 >= 1 + x0 = [[f(_x0)]] We can thus remove the following rules: mark(f(X)) => a!6220!6220f(mark(X)) mark(a) => a mark(c(X)) => c(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(a)) >? c(f(g(f(a)))) mark(g(X)) >? g(mark(X)) a!6220!6220f(X) >? f(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 3 a!6220!6220f = \y0.3 + 3y0 c = \y0.y0 f = \y0.1 + 3y0 g = \y0.y0 mark = \y0.3y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(f(a))]] = 33 > 31 = [[c(f(g(f(a))))]] [[mark(g(_x0))]] = 3x0 >= 3x0 = [[g(mark(_x0))]] [[a!6220!6220f(_x0)]] = 3 + 3x0 > 1 + 3x0 = [[f(_x0)]] We can thus remove the following rules: a!6220!6220f(f(a)) => c(f(g(f(a)))) a!6220!6220f(X) => f(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]): mark(g(X)) >? g(mark(X)) We orient these requirements with a polynomial interpretation in the natural numbers.
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