details

property	value
status	complete
benchmark	Applicative_05__TreeHeight.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.0934979 seconds
cpu usage	0.093642
user time	0.077026
system time	0.016616
max virtual memory	113188.0
max residence set size	3216.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> d cons : [d * c] --> c false : [] --> a height : [] --> d -> d if : [a * d] --> d le : [d * d] --> a map : [d -> d * c] --> c maxlist : [d * c] --> d nil : [] --> c node : [b * c] --> d s : [d] --> d true : [] --> a Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) maxlist(x, cons(y, z)) => if(le(x, y), maxlist(y, z)) maxlist(x, nil) => x height node(x, y) => s(maxlist(0, map(height, y))) 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]): map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) le(0, X) >? true le(s(X), 0) >? false le(s(X), s(Y)) >? le(X, Y) maxlist(X, cons(Y, Z)) >? if(le(X, Y), maxlist(Y, Z)) maxlist(X, nil) >? X height node(X, Y) >? s(maxlist(0, map(height, Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.2 + 2y0 + 2y1 false = 0 height = \y0.0 if = \y0y1.y0 + y1 le = \y0y1.1 + y0 + y1 map = \G0y1.2y1 + G0(0) + 2y1G0(y1) maxlist = \y0y1.1 + y0 + y1 nil = 0 node = \y0y1.3 + y0 + 3y1 s = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[map(_F0, nil)]] = F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 4 + 4x1 + 4x2 + F0(0) + 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 4F0(2 + 2x1 + 2x2) > 2 + 2x1 + 4x2 + 2F0(0) + 2F0(x1) + 4x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[le(0, _x0)]] = 1 + x0 > 0 = [[true]] [[le(s(_x0), 0)]] = 1 + x0 > 0 = [[false]] [[le(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[le(_x0, _x1)]] [[maxlist(_x0, cons(_x1, _x2))]] = 3 + x0 + 2x1 + 2x2 > 2 + x0 + x2 + 2x1 = [[if(le(_x0, _x1), maxlist(_x1, _x2))]] [[maxlist(_x0, nil)]] = 1 + x0 > x0 = [[_x0]] [[height node(_x0, _x1)]] = 3 + x0 + 3x1 > 1 + 2x1 = [[s(maxlist(0, map(height, _x1)))]] We can thus remove the following rules: map(F, cons(X, Y)) => cons(F X, map(F, Y)) le(0, X) => true le(s(X), 0) => false maxlist(X, cons(Y, Z)) => if(le(X, Y), maxlist(Y, Z)) maxlist(X, nil) => X height node(X, Y) => s(maxlist(0, map(height, 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]): map(F, nil) >? nil le(s(X), s(Y)) >? le(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: le = \y0y1.y0 + y1 map = \G0y1.3 + 3y1 + G0(0) nil = 0 s = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] popout

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all output return to Higher Order Rewriting Union Beta