details

property	value
status	complete
benchmark	h58.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)	14.6858 seconds
cpu usage	29.2566
user time	28.7769
system time	0.47971
max virtual memory	270100.0
max residence set size	154188.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat cons : [nat * list] --> list foldr : [nat -> nat -> nat * nat * list] --> nat length : [list] --> nat nil : [] --> list s : [nat] --> nat succ : [] --> nat -> nat -> nat xap : [nat -> nat -> nat * nat] --> nat -> nat yap : [nat -> nat * nat] --> nat 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)) succ x y => s(y) length(x) => foldr(/\y./\z.yap(xap(succ, y), z), 0, x) 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 : [] --> nat cons : [nat * list] --> list foldr : [nat -> nat -> nat * nat * list] --> nat length : [list] --> nat nil : [] --> list s : [nat] --> nat succ : [nat] --> nat -> nat yap : [nat -> nat * nat] --> nat 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)) succ(X) Y => s(Y) length(X) => foldr(/\x./\y.yap(succ(x), y), 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]): 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)) succ(X) Y >? s(Y) length(X) >? foldr(/\x./\y.yap(succ(x), y), 0, X) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[s(x_1)]] = x_1 We choose Lex = {} and Mul = {@_{o -> o}, cons, foldr, length, nil, succ, yap}, and the following precedence: cons > length > foldr > nil > succ > yap > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 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(/\x./\y.yap(F(x), y), X, Z)) @_{o -> o}(succ(X), Y) > Y length(X) >= foldr(/\x./\y.yap(succ(x), y), _|_, X) yap(F, X) > @_{o -> o}(F, X) With these choices, we have: 1] foldr(/\x./\y.yap(F(x), y), X, nil) >= X because [2], by (Star) 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) popout

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all output return to Higher Order Rewriting Union Beta