details

property	value
status	complete
benchmark	AotoYamada_05__011.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	0.112243890762 seconds
cpu usage	0.109291933
max memory	5824512.0

stage attributes

key	value
output-size	3565
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [b * a] --> a curry : [b -> b -> b * b] --> b -> b inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [] --> b -> b -> b s : [b] --> b Rules: plus 0 x => x plus s(x) y => s(plus x y) map(f) nil => nil map(f) cons(x, y) => cons(f x, map(f) y) curry(f, x) y => f x y inc => map(curry(plus, s(0))) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): plus 0 X >? X plus s(X) Y >? s(plus X Y) map(F) nil >? nil map(F) cons(X, Y) >? cons(F X, map(F) Y) curry(F, X) Y >? F X Y inc >? map(curry(plus, s(0))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y0 + y1 curry = \G0y1y2.y1 + 2G0(y1,y2) inc = \y0.3 + 3y0 map = \G0y1.1 + G0(y1) + 2y1G0(y1) nil = 3 plus = \y0y1.0 s = \y0.y0 Using this interpretation, the requirements translate to: [[plus 0 _x0]] = x0 >= x0 = [[_x0]] [[plus s(_x0) _x1]] = x0 + x1 >= x0 + x1 = [[s(plus _x0 _x1)]] [[map(_F0) nil]] = 4 + 7F0(3) > 3 = [[nil]] [[map(_F0) cons(_x1, _x2)]] = 4 + x1 + x2 + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 7F0(3 + x1 + x2) >= 4 + x1 + x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0) _x2)]] [[curry(_F0, _x1) _x2]] = x1 + x2 + 2F0(x1,x2) >= x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] [[inc]] = \y0.3 + 3y0 > \y0.1 = [[map(curry(plus, s(0)))]] We can thus remove the following rules: map(F) nil => nil inc => map(curry(plus, s(0))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) map(F, cons(X, Y)) >? cons(F X, map(F, Y)) curry(F, X, Y) >? F X Y We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 cons = \y0y1.3 + y0 + y1 curry = \G0y1y2.3 + y1 + y2 + G0(y1,y2) map = \G0y1.3y1 + G0(0) + 2y1G0(y1) plus = \y0y1.3 + y1 + 3y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[plus(0, _x0)]] = 12 + x0 > x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + F0(0) + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 6F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[curry(_F0, _x1, _x2)]] = 3 + x1 + x2 + F0(x1,x2) > x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] We can thus remove the following rules: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) popout

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job log

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actions

all output return to HRS union beta 16688