details

property	value
status	complete
benchmark	h13.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	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	5.76431 seconds
cpu usage	5.76305
user time	5.66856
system time	0.094488
max virtual memory	187260.0
max residence set size	70788.0

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 xap : [a -> b -> b * a] --> b -> b yap : [b -> b * b] --> b Rules: rec(/\x./\y.yap(xap(f, x), y), z, 0) => z rec(/\x./\y.yap(xap(f, x), y), z, s(u)) => yap(xap(f, s(u)), rec(/\v./\w.yap(xap(f, v), w), z, u)) xap(f, x) => f x yap(f, x) => f x This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). Symbol xap 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 yap : [b -> b * b] --> b Rules: rec(/\x./\y.yap(F(x), y), X, 0) => X rec(/\x./\y.yap(F(x), y), X, s(Y)) => yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) yap(F, X) => F X 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.yap(F(x), y), X, 0) >? X rec(/\x./\y.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {0, @_{o -> o}, rec, s, yap}, and the following precedence: 0 > rec > s > yap > @_{o -> o} With these choices, we have: 1] rec(/\x./\y.yap(F(x), y), X, 0) > X because [2], by definition 2] rec*(/\x./\y.yap(F(x), y), X, 0) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y)) because [5], by (Star) 5] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y)) because rec > yap, [6] and [14], by (Copy) 6] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= F(s(Y)) because [7], by (Select) 7] /\x.yap(F(rec*(/\y./\z.yap(F(y), z), X, s(Y))), x) >= F(s(Y)) because [8], by (Eta)[Kop13:2] 8] F(rec*(/\x./\y.yap(F(x), y), X, s(Y))) >= F(s(Y)) because [9], by (Meta) 9] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= s(Y) because rec > s and [10], by (Copy) 10] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= Y because [11], by (Select) 11] s(Y) >= Y because [12], by (Star) 12] s*(Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= rec(/\x./\y.yap(F(x), y), X, Y) because rec in Mul, [15], [21] and [22], by (Stat) 15] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [16], by (Abs) 16] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [17], by (Abs) 17] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [18] and [20], by (Fun) 18] F(y) >= F(y) because [19], by (Meta) 19] y >= y by (Var) 20] x >= x by (Var) 21] X >= X by (Meta) 22] s(Y) > Y because [23], by definition 23] s*(Y) >= Y because [13], by (Select) 24] yap(F, X) > @_{o -> o}(F, X) because [25], by definition 25] yap*(F, X) >= @_{o -> o}(F, X) because yap > @_{o -> o}, [26] and [28], by (Copy) 26] yap*(F, X) >= F because [27], by (Select) 27] F >= F by (Meta) 28] yap*(F, X) >= X because [29], by (Select) 29] X >= X by (Meta) We can thus remove the following rules: rec(/\x./\y.yap(F(x), y), X, 0) => X yap(F, X) => F X 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.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: rec = \G0y1y2.y1 + y2 + G0(0,0) + 2y2y2G0(y2,y2) + 2G0(y2,y1) s = \y0.3 + 3y0 yap = \G0y1.y1 + 2G0(0) popout

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

job log

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actions

all output return to Higher Order Rewriting Union Beta