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TRS Stand 20472 pair #381715870
details
property
value
status
complete
benchmark
Ex1_Zan97_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n099.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0477797985077 seconds
cpu usage
0.029680843
max memory
1384448.0
stage attributes
key
value
output-size
3320
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 a!6220!6220g : [o] --> o a!6220!6220h : [o] --> o c : [] --> o d : [] --> o g : [o] --> o h : [o] --> o mark : [o] --> o a!6220!6220g(X) => a!6220!6220h(X) a!6220!6220c => d a!6220!6220h(d) => a!6220!6220g(c) mark(g(X)) => a!6220!6220g(X) mark(h(X)) => a!6220!6220h(X) mark(c) => a!6220!6220c mark(d) => d a!6220!6220g(X) => g(X) a!6220!6220h(X) => h(X) a!6220!6220c => c 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!6220g(X) >? a!6220!6220h(X) a!6220!6220c >? d a!6220!6220h(d) >? a!6220!6220g(c) mark(g(X)) >? a!6220!6220g(X) mark(h(X)) >? a!6220!6220h(X) mark(c) >? a!6220!6220c mark(d) >? d a!6220!6220g(X) >? g(X) a!6220!6220h(X) >? h(X) a!6220!6220c >? c We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220c = 1 a!6220!6220g = \y0.1 + 2y0 a!6220!6220h = \y0.1 + y0 c = 0 d = 0 g = \y0.y0 h = \y0.y0 mark = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[a!6220!6220g(_x0)]] = 1 + 2x0 >= 1 + x0 = [[a!6220!6220h(_x0)]] [[a!6220!6220c]] = 1 > 0 = [[d]] [[a!6220!6220h(d)]] = 1 >= 1 = [[a!6220!6220g(c)]] [[mark(g(_x0))]] = 3 + 3x0 > 1 + 2x0 = [[a!6220!6220g(_x0)]] [[mark(h(_x0))]] = 3 + 3x0 > 1 + x0 = [[a!6220!6220h(_x0)]] [[mark(c)]] = 3 > 1 = [[a!6220!6220c]] [[mark(d)]] = 3 > 0 = [[d]] [[a!6220!6220g(_x0)]] = 1 + 2x0 > x0 = [[g(_x0)]] [[a!6220!6220h(_x0)]] = 1 + x0 > x0 = [[h(_x0)]] [[a!6220!6220c]] = 1 > 0 = [[c]] We can thus remove the following rules: a!6220!6220c => d mark(g(X)) => a!6220!6220g(X) mark(h(X)) => a!6220!6220h(X) mark(c) => a!6220!6220c mark(d) => d a!6220!6220g(X) => g(X) a!6220!6220h(X) => h(X) a!6220!6220c => c 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!6220g(X) >? a!6220!6220h(X) a!6220!6220h(d) >? a!6220!6220g(c) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220g = \y0.2 + 3y0 a!6220!6220h = \y0.3y0 c = 0 d = 3 Using this interpretation, the requirements translate to:
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