details

property	value
status	complete
benchmark	Applicative_05__TreeFlatten.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.148261 seconds
cpu usage	0.148517
user time	0.127657
system time	0.02086
max virtual memory	113188.0
max residence set size	3436.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: append : [a * a] --> a concat : [a] --> a cons : [a * a] --> a flatten : [] --> a -> a map : [a -> a * a] --> a nil : [] --> a node : [a * a] --> a Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) flatten node(x, y) => cons(x, concat(map(flatten, y))) concat(nil) => nil concat(cons(x, y)) => append(x, concat(y)) append(nil, x) => x append(cons(x, y), z) => cons(x, append(y, z)) 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)) flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) concat(nil) >? nil concat(cons(X, Y)) >? append(X, concat(Y)) append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.y0 + y1 concat = \y0.y0 cons = \y0y1.1 + y0 + y1 flatten = \y0.0 map = \G0y1.2 + y1 + G0(y1) + 2y1G0(y1) nil = 0 node = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) >= 3 + x1 + x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + x1 = [[cons(_x0, concat(map(flatten, _x1)))]] [[concat(nil)]] = 0 >= 0 = [[nil]] [[concat(cons(_x0, _x1))]] = 1 + x0 + x1 > x0 + x1 = [[append(_x0, concat(_x1))]] [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] We can thus remove the following rules: map(F, nil) => nil concat(cons(X, Y)) => append(X, concat(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, cons(X, Y)) >? cons(F X, map(F, Y)) flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) concat(nil) >? nil append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.3 + y1 + 3y0 concat = \y0.y0 cons = \y0y1.3 + y0 + y1 flatten = \y0.0 map = \G0y1.2y1 + G0(0) + 3y1G0(y1) nil = 0 node = \y0y1.3 + 3y0 + 3y1 Using this interpretation, the requirements translate to: [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) > 3 + x1 + 2x2 + F0(0) + F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + 2x1 = [[cons(_x0, concat(map(flatten, _x1)))]] [[concat(nil)]] = 0 >= 0 = [[nil]] [[append(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] We can thus remove the following rules: map(F, cons(X, Y)) => cons(F X, map(F, Y)) append(nil, X) => X append(cons(X, Y), Z) => cons(X, append(Y, Z)) popout

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actions

all output return to Higher Order Rewriting Union Beta