details

property	value
status	complete
benchmark	length.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.166148 seconds
cpu usage	0.166403
user time	0.142652
system time	0.023751
max virtual memory	113188.0
max residence set size	5728.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat cons : [nat * list] --> list foldr : [nat -> nat -> nat * nat * list] --> nat length : [list] --> nat nil : [] --> list s : [nat] --> nat Rules: foldr(f, x, nil) => x foldr(f, x, cons(y, z)) => f y foldr(f, x, z) length(x) => foldr(/\y./\z.s(z), 0, x) 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]): foldr(F, X, nil) >? X foldr(F, X, cons(Y, Z)) >? F Y foldr(F, X, Z) length(X) >? foldr(/\x./\y.s(y), 0, X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[s(x_1)]] = x_1 We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, foldr, length, nil}, and the following precedence: length > foldr > @_{o -> o} > @_{o -> o -> o} > cons > nil Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldr(F, X, nil) >= X foldr(F, X, cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) length(X) >= foldr(/\x./\y.y, _|_, X) With these choices, we have: 1] foldr(F, X, nil) >= X because [2], by (Star) 2] foldr*(F, X, nil) >= X because [3], by (Select) 3] X >= X by (Meta) 4] foldr(F, X, cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) because [5], by definition 5] foldr*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) because foldr > @_{o -> o}, [6] and [13], by (Copy) 6] foldr*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, Y) because foldr > @_{o -> o -> o}, [7] and [9], by (Copy) 7] foldr*(F, X, cons(Y, Z)) >= F because [8], by (Select) 8] F >= F by (Meta) 9] foldr*(F, X, cons(Y, Z)) >= Y because [10], by (Select) 10] cons(Y, Z) >= Y because [11], by (Star) 11] cons*(Y, Z) >= Y because [12], by (Select) 12] Y >= Y by (Meta) 13] foldr*(F, X, cons(Y, Z)) >= foldr(F, X, Z) because foldr in Mul, [14], [15] and [16], by (Stat) 14] F >= F by (Meta) 15] X >= X by (Meta) 16] cons(Y, Z) > Z because [17], by definition 17] cons*(Y, Z) >= Z because [18], by (Select) 18] Z >= Z by (Meta) 19] length(X) >= foldr(/\x./\y.y, _|_, X) because [20], by (Star) 20] length*(X) >= foldr(/\x./\y.y, _|_, X) because length > foldr, [21], [25] and [26], by (Copy) 21] length*(X) >= /\y./\z.z because [22], by (F-Abs) 22] length*(X, x) >= /\z.z because [23], by (F-Abs) 23] length*(X, x, y) >= y because [24], by (Select) 24] y >= y by (Var) 25] length*(X) >= _|_ by (Bot) 26] length*(X) >= X because [27], by (Select) 27] X >= X by (Meta) We can thus remove the following rules: foldr(F, X, cons(Y, Z)) => F Y foldr(F, X, 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]): foldr(F, X, nil) >? X length(X) >? foldr(/\x./\y.s(y), 0, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 foldr = \G0y1y2.y1 + y2 + G0(0,0) length = \y0.3 + 3y0 nil = 3 s = \y0.y0 Using this interpretation, the requirements translate to: [[foldr(_F0, _x1, nil)]] = 3 + x1 + F0(0,0) > x1 = [[_x1]] [[length(_x0)]] = 3 + 3x0 > x0 = [[foldr(/\x./\y.s(y), 0, _x0)]] popout

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all output return to Higher Order Rewriting Union Beta