details

property	value
status	complete
benchmark	Applicative_AG01_innermost__#4.28.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.363873 seconds
cpu usage	0.350447
user time	0.308063
system time	0.042384
max virtual memory	113188.0
max residence set size	6072.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> a bits : [a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d half : [a] --> a map : [c -> c * d] --> d nil : [] --> d s : [a] --> a true : [] --> b Rules: half(0) => 0 half(s(0)) => 0 half(s(s(x))) => s(half(x)) bits(0) => 0 bits(s(x)) => s(bits(half(s(x)))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, 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]): half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) bits(0) >? 0 bits(s(X)) >? s(bits(half(s(X)))) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) filter2(false, F, X, Y) >? filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 bits = \y0.y0 cons = \y0y1.3 + y1 + 2y0 false = 3 filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) filter2 = \y0G1y2y3.2y0 + 2y2 + 2y3 + 2G1(0) + 3y3G1(y3) half = \y0.y0 map = \G0y1.2 + 3y1 + G0(y1) + 2y1G0(y1) nil = 1 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[half(0)]] = 0 >= 0 = [[0]] [[half(s(0))]] = 0 >= 0 = [[0]] [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] [[bits(0)]] = 0 >= 0 = [[0]] [[bits(s(_x0))]] = x0 >= x0 = [[s(bits(half(s(_x0))))]] [[map(_F0, nil)]] = 5 + 3F0(1) > 1 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 11 + 3x2 + 6x1 + 2x2F0(3 + x2 + 2x1) + 4x1F0(3 + x2 + 2x1) + 7F0(3 + x2 + 2x1) > 5 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2 + F0(0) + 3F0(1) > 1 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) > 2x2 + 4x1 + 2F0(0) + 2F0(x1) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 3 + 2x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 2x2 + F0(0) + 3x2F0(x2) = [[filter(_F0, _x2)]] We can thus remove the following rules: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, 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]): half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) bits(0) >? 0 bits(s(X)) >? s(bits(half(s(X)))) We orient these requirements with a polynomial interpretation in the natural numbers. popout

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all output return to Higher Order Rewriting Union Beta