details

property	value
status	complete
benchmark	h59.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.115857 seconds
cpu usage	0.097161
user time	0.077162
system time	0.019999
max virtual memory	113188.0
max residence set size	4784.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat rec : [] --> nat -> a -> (nat -> a -> a) -> a s : [] --> nat -> nat Rules: rec 0 x (/\y./\z.f y z) => x rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) 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 : [] --> nat rec : [nat * a * nat -> a -> a] --> a s : [nat] --> nat ~AP1 : [nat -> a -> a * nat] --> a -> a Rules: rec(0, X, /\x./\y.~AP1(F, x) y) => X rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) ~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 : [] --> nat rec : [nat * a * nat -> a -> a] --> a s : [nat] --> nat Rules: rec(0, X, /\x./\y.Y(x, y)) => X rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 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(0, X, /\x./\y.Y(x, y)) >? X rec(s(X), Y, /\x./\y.Z(x, y)) >? Z(X, rec(X, Y, /\z./\u.Z(z, u))) 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 > rec > s With these choices, we have: 1] rec(0, X, /\x./\y.Y(x, y)) >= X because [2], by (Star) 2] rec*(0, X, /\x./\y.Y(x, y)) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, /\x./\y.Z(x, y)) > Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [5], by definition 5] rec*(s(X), Y, /\x./\y.Z(x, y)) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [6], by (Select) 6] Z(rec*(s(X), Y, /\x./\y.Z(x, y)), rec*(s(X), Y, /\z./\u.Z(z, u))) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [7] and [11], by (Meta) 7] rec*(s(X), Y, /\x./\y.Z(x, y)) >= X because [8], by (Select) 8] s(X) >= X because [9], by (Star) 9] s*(X) >= X because [10], by (Select) 10] X >= X by (Meta) 11] rec*(s(X), Y, /\x./\y.Z(x, y)) >= rec(X, Y, /\x./\y.Z(x, y)) because rec in Mul, [12], [14] and [15], by (Stat) 12] s(X) > X because [13], by definition 13] s*(X) >= X because [10], by (Select) 14] Y >= Y by (Meta) 15] /\x./\z.Z(x, z) >= /\x./\z.Z(x, z) because [16], by (Abs) 16] /\z.Z(y, z) >= /\z.Z(y, z) because [17], by (Abs) 17] Z(y, x) >= Z(y, x) because [18] and [19], by (Meta) 18] y >= y by (Var) 19] x >= x by (Var) We can thus remove the following rules: rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 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(0, X, /\x./\y.Y(x, y)) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 rec = \y0y1G2.3 + y0 + y1 + G2(0,0) Using this interpretation, the requirements translate to: [[rec(0, _x0, /\x./\y._x1(x, y))]] = 6 + x0 + F1(0,0) > x0 = [[_x0]] 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