details

property	value
status	complete
benchmark	h12.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.101363 seconds
cpu usage	0.101649
user time	0.081241
system time	0.020408
max virtual memory	113188.0
max residence set size	4764.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> a rec : [] --> (a -> b -> b) -> b -> a -> b s : [] --> a -> a Rules: rec (/\x./\y.f x y) z 0 => z rec (/\x./\y.f x y) z (s u) => f (s u) (rec (/\v./\w.f v w) z u) Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: Alphabet: 0 : [] --> a rec : [a -> b -> b * b * a] --> b s : [a] --> a ~AP1 : [a -> b -> b * a] --> b -> b Rules: rec(/\x./\y.~AP1(F, x) y, X, 0) => X rec(/\x./\y.~AP1(F, x) y, X, s(Y)) => ~AP1(F, s(Y)) rec(/\z./\u.~AP1(F, z) u, X, Y) ~AP1(F, X) => F X Symbol ~AP1 is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. This gives: Alphabet: 0 : [] --> a rec : [a -> b -> b * b * a] --> b s : [a] --> a Rules: rec(/\x./\y.X(x, y), Y, 0) => Y rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, 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]): rec(/\x./\y.X(x, y), Y, 0) >? Y rec(/\x./\y.X(x, y), Y, s(Z)) >? X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {0, rec, s}, and the following precedence: 0 > s > rec With these choices, we have: 1] rec(/\x./\y.X(x, y), Y, 0) >= Y because [2], by (Star) 2] rec*(/\x./\y.X(x, y), Y, 0) >= Y because [3], by (Select) 3] Y >= Y by (Meta) 4] rec(/\x./\y.X(x, y), Y, s(Z)) > X(s(Z), rec(/\x./\y.X(x, y), Y, Z)) because [5], by definition 5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(s(Z), rec(/\x./\y.X(x, y), Y, Z)) because [6], by (Select) 6] X(rec*(/\x./\y.X(x, y), Y, s(Z)), rec*(/\z./\u.X(z, u), Y, s(Z))) >= X(s(Z), rec(/\x./\y.X(x, y), Y, Z)) because [7] and [10], by (Meta) 7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= s(Z) because [8], by (Select) 8] s(Z) >= s(Z) because s in Mul and [9], by (Fun) 9] Z >= Z by (Meta) 10] rec*(/\x./\y.X(x, y), Y, s(Z)) >= rec(/\x./\y.X(x, y), Y, Z) because rec in Mul, [11], [16] and [17], by (Stat) 11] /\x./\z.X(x, z) >= /\x./\z.X(x, z) because [12], by (Abs) 12] /\z.X(y, z) >= /\z.X(y, z) because [13], by (Abs) 13] X(y, x) >= X(y, x) because [14] and [15], by (Meta) 14] y >= y by (Var) 15] x >= x by (Var) 16] Y >= Y by (Meta) 17] s(Z) > Z because [18], by definition 18] s*(Z) >= Z because [9], by (Select) We can thus remove the following rules: rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, 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]): rec(/\x./\y.X(x, y), Y, 0) >? Y We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 rec = \G0y1y2.3 + y1 + y2 + G0(0,0) Using this interpretation, the requirements translate to: [[rec(/\x./\y._x0(x, y), _x1, 0)]] = 6 + x1 + F0(0,0) > x1 = [[_x1]] 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