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TRS Stand 20472 pair #381715205
details
property
value
status
complete
benchmark
ExProp7_Luc06_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n055.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0819399356842 seconds
cpu usage
0.078573736
max memory
2342912.0
stage attributes
key
value
output-size
7423
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. 0 : [] --> o activate : [o] --> o cons : [o * o] --> o f : [o] --> o n!6220!62200 : [] --> o n!6220!6220f : [o] --> o n!6220!6220s : [o] --> o p : [o] --> o s : [o] --> o f(0) => cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) f(s(0)) => f(p(s(0))) p(s(X)) => X f(X) => n!6220!6220f(X) s(X) => n!6220!6220s(X) 0 => n!6220!62200 activate(n!6220!6220f(X)) => f(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!62200) => 0 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(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) f(s(0)) >? f(p(s(0))) p(s(X)) >? X f(X) >? n!6220!6220f(X) s(X) >? n!6220!6220s(X) 0 >? n!6220!62200 activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220s(X)) >? s(activate(X)) activate(n!6220!62200) >? 0 activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 activate = \y0.1 + 2y0 cons = \y0y1.y0 + y1 f = \y0.y0 n!6220!62200 = 0 n!6220!6220f = \y0.y0 n!6220!6220s = \y0.y0 p = \y0.y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[f(0)]] = 0 >= 0 = [[cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200)))]] [[f(s(0))]] = 0 >= 0 = [[f(p(s(0)))]] [[p(s(_x0))]] = x0 >= x0 = [[_x0]] [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] [[s(_x0)]] = x0 >= x0 = [[n!6220!6220s(_x0)]] [[0]] = 0 >= 0 = [[n!6220!62200]] [[activate(n!6220!6220f(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[f(activate(_x0))]] [[activate(n!6220!6220s(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[s(activate(_x0))]] [[activate(n!6220!62200)]] = 1 > 0 = [[0]] [[activate(_x0)]] = 1 + 2x0 > x0 = [[_x0]] We can thus remove the following rules: activate(n!6220!62200) => 0 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(0) >? cons(0, n!6220!6220f(n!6220!6220s(n!6220!62200))) f(s(0)) >? f(p(s(0))) p(s(X)) >? X f(X) >? n!6220!6220f(X) s(X) >? n!6220!6220s(X) 0 >? n!6220!62200 activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220s(X)) >? s(activate(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 activate = \y0.y0 cons = \y0y1.y0 + y1 f = \y0.y0
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