details

property	value
status	complete
benchmark	iterative.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	1.56208 seconds
cpu usage	1.56239
user time	1.49387
system time	0.068527
max virtual memory	145704.0
max residence set size	30080.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: a : [] --> o b : [] --> o f : [o * o -> o] --> o g : [o * o * o -> o] --> o Rules: g(x, y, h) => f(f(x, h), h) f(x, h) => b b => a f(b, /\x.g(x, x, h)) => g(f(a, /\y.g(y, y, h)), f(b, /\z.b), h) 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]): g(X, Y, F) >? f(f(X, F), F) f(X, F) >? b b >? a f(b, /\x.g(x, x, F)) >? g(f(a, /\y.g(y, y, F)), f(b, /\z.b), F) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[a]] = _|_ We choose Lex = {} and Mul = {b, f, g}, and the following precedence: g > f > b Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: g(X, Y, F) >= f(f(X, F), F) f(X, F) >= b b > _|_ f(b, /\x.g(x, x, F)) > g(f(_|_, /\x.g(x, x, F)), f(b, /\y.b), F) With these choices, we have: 1] g(X, Y, F) >= f(f(X, F), F) because [2], by (Star) 2] g*(X, Y, F) >= f(f(X, F), F) because g > f, [3] and [6], by (Copy) 3] g*(X, Y, F) >= f(X, F) because g > f, [4] and [6], by (Copy) 4] g*(X, Y, F) >= X because [5], by (Select) 5] X >= X by (Meta) 6] g*(X, Y, F) >= F because [7], by (Select) 7] F >= F by (Meta) 8] f(X, F) >= b because [9], by (Star) 9] f*(X, F) >= b because f > b, by (Copy) 10] b > _|_ because [11], by definition 11] b* >= _|_ by (Bot) 12] f(b, /\x.g(x, x, F)) > g(f(_|_, /\x.g(x, x, F)), f(b, /\y.b), F) because [13], by definition 13] f*(b, /\x.g(x, x, F)) >= g(f(_|_, /\x.g(x, x, F)), f(b, /\y.b), F) because [14], by (Select) 14] g(f*(b, /\x.g(x, x, F)), f*(b, /\y.g(y, y, F)), F) >= g(f(_|_, /\x.g(x, x, F)), f(b, /\y.b), F) because g in Mul, [15], [20] and [19], by (Fun) 15] f*(b, /\x.g(x, x, F)) >= f(_|_, /\x.g(x, x, F)) because f in Mul, [10] and [16], by (Stat) 16] /\y.g(y, y, F) >= /\y.g(y, y, F) because [17], by (Abs) 17] g(x, x, F) >= g(x, x, F) because g in Mul, [18], [18] and [19], by (Fun) 18] x >= x by (Var) 19] F >= F by (Meta) 20] f*(b, /\y.g(y, y, F)) >= f(b, /\y.b) because [21], by (Select) 21] g(f*(b, /\y.g(y, y, F)), f*(b, /\z.g(z, z, F)), F) >= f(b, /\y.b) because [22], by (Star) 22] g*(f*(b, /\y.g(y, y, F)), f*(b, /\z.g(z, z, F)), F) >= f(b, /\y.b) because g > f, [23] and [24], by (Copy) 23] g*(f*(b, /\y.g(y, y, F)), f*(b, /\z.g(z, z, F)), F) >= b because g > b, by (Copy) 24] g*(f*(b, /\y.g(y, y, F)), f*(b, /\z.g(z, z, F)), F) >= /\y.b because [25], by (F-Abs) 25] g*(f*(b, /\y.g(y, y, F)), f*(b, /\z.g(z, z, F)), F, u) >= b because g > b, by (Copy) We can thus remove the following rules: b => a f(b, /\x.g(x, x, F)) => g(f(a, /\y.g(y, y, F)), f(b, /\z.b), F) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): g(X, Y, F) >? f(f(X, F), F) f(X, F) >? b We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: b = 0 f = \y0G1.y0 + G1(0) g = \y0y1G2.3 + y1 + 3y0 + G2(y0) + 2G2(0) + 2G2(y1) Using this interpretation, the requirements translate to: [[g(_x0, _x1, _F2)]] = 3 + x1 + 3x0 + F2(x0) + 2F2(0) + 2F2(x1) > x0 + 2F2(0) = [[f(f(_x0, _F2), _F2)]] [[f(_x0, _F1)]] = x0 + F1(0) >= 0 = [[b]] We can thus remove the following rules: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Higher Order Rewriting Union Beta