details

property	value
status	complete
benchmark	h56.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	7.20951 seconds
cpu usage	7.20829
user time	7.09723
system time	0.11106
max virtual memory	186848.0
max residence set size	70200.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: cons : [a * b] --> b foldr : [a -> b -> b * b * b] --> b nil : [] --> b xap : [a -> b -> b * a] --> b -> b yap : [b -> b * b] --> b Rules: foldr(/\x./\y.yap(xap(f, x), y), z, nil) => z foldr(/\x./\y.yap(xap(f, x), y), z, cons(u, v)) => yap(xap(f, u), foldr(/\w./\x'.yap(xap(f, w), x'), z, v)) 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: cons : [a * b] --> b foldr : [a -> b -> b * b * b] --> b nil : [] --> b yap : [b -> b * b] --> b Rules: foldr(/\x./\y.yap(F(x), y), X, nil) => X foldr(/\x./\y.yap(F(x), y), X, cons(Y, Z)) => yap(F(Y), foldr(/\z./\u.yap(F(z), u), X, Z)) 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]): foldr(/\x./\y.yap(F(x), y), X, nil) >? X foldr(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? yap(F(Y), foldr(/\z./\u.yap(F(z), u), X, Z)) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o}, cons, foldr, nil, yap}, and the following precedence: foldr > nil > yap > @_{o -> o} > cons With these choices, we have: 1] foldr(/\x./\y.yap(F(x), y), X, nil) > X because [2], by definition 2] foldr*(/\x./\y.yap(F(x), y), X, nil) >= X because [3], by (Select) 3] X >= X by (Meta) 4] foldr(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= yap(F(Y), foldr(/\x./\y.yap(F(x), y), X, Z)) because [5], by (Star) 5] foldr*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= yap(F(Y), foldr(/\x./\y.yap(F(x), y), X, Z)) because foldr > yap, [6] and [13], by (Copy) 6] foldr*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= F(Y) because [7], by (Select) 7] /\x.yap(F(foldr*(/\y./\z.yap(F(y), z), X, cons(Y, Z))), x) >= F(Y) because [8], by (Eta)[Kop13:2] 8] F(foldr*(/\x./\y.yap(F(x), y), X, cons(Y, Z))) >= F(Y) because [9], by (Meta) 9] foldr*(/\x./\y.yap(F(x), y), 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*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldr(/\x./\y.yap(F(x), y), X, Z) because foldr in Mul, [14], [20] and [21], by (Stat) 14] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [15], by (Abs) 15] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [16], by (Abs) 16] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [17] and [19], by (Fun) 17] F(y) >= F(y) because [18], by (Meta) 18] y >= y by (Var) 19] x >= x by (Var) 20] X >= X by (Meta) 21] cons(Y, Z) > Z because [22], by definition 22] cons*(Y, Z) >= Z because [23], by (Select) 23] Z >= Z by (Meta) 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: foldr(/\x./\y.yap(F(x), y), X, nil) => 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]): foldr(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? yap(F(Y), foldr(/\z./\u.yap(F(z), u), X, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.3 + y0 + 3y1 foldr = \G0y1y2.y1 + 3y2 + G0(y2,y1) + 3y2y2G0(y2,y2) + y1y2G0(y1,y2) yap = \G0y1.2y1 + G0(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