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TRS Stand 20472 pair #381715064
details
property
value
status
complete
benchmark
Ex5_Zan97_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n052.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0620460510254 seconds
cpu usage
0.045400333
max memory
1462272.0
stage attributes
key
value
output-size
4704
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. activate : [o] --> o c : [] --> o f : [o] --> o false : [] --> o if : [o * o * o] --> o n!6220!6220f : [o] --> o n!6220!6220true : [] --> o true : [] --> o f(X) => if(X, c, n!6220!6220f(n!6220!6220true)) if(true, X, Y) => X if(false, X, Y) => activate(Y) f(X) => n!6220!6220f(X) true => n!6220!6220true activate(n!6220!6220f(X)) => f(activate(X)) activate(n!6220!6220true) => true 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) >? if(X, c, n!6220!6220f(n!6220!6220true)) if(true, X, Y) >? X if(false, X, Y) >? activate(Y) f(X) >? n!6220!6220f(X) true >? n!6220!6220true activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220true) >? true activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.2y0 c = 0 f = \y0.y0 false = 3 if = \y0y1y2.y0 + y1 + 2y2 n!6220!6220f = \y0.y0 n!6220!6220true = 0 true = 0 Using this interpretation, the requirements translate to: [[f(_x0)]] = x0 >= x0 = [[if(_x0, c, n!6220!6220f(n!6220!6220true))]] [[if(true, _x0, _x1)]] = x0 + 2x1 >= x0 = [[_x0]] [[if(false, _x0, _x1)]] = 3 + x0 + 2x1 > 2x1 = [[activate(_x1)]] [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] [[true]] = 0 >= 0 = [[n!6220!6220true]] [[activate(n!6220!6220f(_x0))]] = 2x0 >= 2x0 = [[f(activate(_x0))]] [[activate(n!6220!6220true)]] = 0 >= 0 = [[true]] [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] We can thus remove the following rules: if(false, X, Y) => activate(Y) 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) >? if(X, c, n!6220!6220f(n!6220!6220true)) if(true, X, Y) >? X f(X) >? n!6220!6220f(X) true >? n!6220!6220true activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220true) >? true activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.1 + y0 c = 0 f = \y0.y0 if = \y0y1y2.y0 + y1 + y2 n!6220!6220f = \y0.y0 n!6220!6220true = 0 true = 1 Using this interpretation, the requirements translate to: [[f(_x0)]] = x0 >= x0 = [[if(_x0, c, n!6220!6220f(n!6220!6220true))]] [[if(true, _x0, _x1)]] = 1 + x0 + x1 > x0 = [[_x0]] [[f(_x0)]] = x0 >= x0 = [[n!6220!6220f(_x0)]] [[true]] = 1 > 0 = [[n!6220!6220true]]
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