details

property	value
status	complete
benchmark	Ex5_Zan97_GM.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)	0.109941005707 seconds
cpu usage	0.107301595
max memory	2695168.0

stage attributes

key	value
output-size	8993
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!6220!6220f : [o] --> o a!6220!6220if : [o * o * o] --> o c : [] --> o f : [o] --> o false : [] --> o if : [o * o * o] --> o mark : [o] --> o true : [] --> o a!6220!6220f(X) => a!6220!6220if(mark(X), c, f(true)) a!6220!6220if(true, X, Y) => mark(X) a!6220!6220if(false, X, Y) => mark(Y) mark(f(X)) => a!6220!6220f(mark(X)) mark(if(X, Y, Z)) => a!6220!6220if(mark(X), mark(Y), Z) mark(c) => c mark(true) => true mark(false) => false a!6220!6220f(X) => f(X) a!6220!6220if(X, Y, Z) => if(X, Y, Z) 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(X) >? a!6220!6220if(mark(X), c, f(true)) a!6220!6220if(true, X, Y) >? mark(X) a!6220!6220if(false, X, Y) >? mark(Y) mark(f(X)) >? a!6220!6220f(mark(X)) mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) mark(c) >? c mark(true) >? true mark(false) >? false a!6220!6220f(X) >? f(X) a!6220!6220if(X, Y, Z) >? if(X, Y, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220f = \y0.2 + 2y0 a!6220!6220if = \y0y1y2.y0 + 2y1 + 2y2 c = 0 f = \y0.1 + 2y0 false = 0 if = \y0y1y2.y0 + 2y1 + 2y2 mark = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[a!6220!6220f(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220if(mark(_x0), c, f(true))]] [[a!6220!6220if(true, _x0, _x1)]] = 2x0 + 2x1 >= 2x0 = [[mark(_x0)]] [[a!6220!6220if(false, _x0, _x1)]] = 2x0 + 2x1 >= 2x1 = [[mark(_x1)]] [[mark(f(_x0))]] = 2 + 4x0 >= 2 + 4x0 = [[a!6220!6220f(mark(_x0))]] [[mark(if(_x0, _x1, _x2))]] = 2x0 + 4x1 + 4x2 >= 2x0 + 2x2 + 4x1 = [[a!6220!6220if(mark(_x0), mark(_x1), _x2)]] [[mark(c)]] = 0 >= 0 = [[c]] [[mark(true)]] = 0 >= 0 = [[true]] [[mark(false)]] = 0 >= 0 = [[false]] [[a!6220!6220f(_x0)]] = 2 + 2x0 > 1 + 2x0 = [[f(_x0)]] [[a!6220!6220if(_x0, _x1, _x2)]] = x0 + 2x1 + 2x2 >= x0 + 2x1 + 2x2 = [[if(_x0, _x1, _x2)]] We can thus remove the following rules: 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(X) >? a!6220!6220if(mark(X), c, f(true)) a!6220!6220if(true, X, Y) >? mark(X) a!6220!6220if(false, X, Y) >? mark(Y) mark(f(X)) >? a!6220!6220f(mark(X)) mark(if(X, Y, Z)) >? a!6220!6220if(mark(X), mark(Y), Z) mark(c) >? c mark(true) >? true mark(false) >? false a!6220!6220if(X, Y, Z) >? if(X, Y, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220f = \y0.2y0 a!6220!6220if = \y0y1y2.y0 + 2y1 + 2y2 c = 0 f = \y0.2y0 false = 1 if = \y0y1y2.y0 + y2 + 2y1 popout

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