details

property	value
status	complete
benchmark	h60.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	9.3104 seconds
cpu usage	9.29387
user time	9.10643
system time	0.18744
max virtual memory	397268.0
max residence set size	115860.0

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 xap : [nat -> a -> a * nat] --> a -> a yap : [a -> a * a] --> a Rules: rec(0, x, /\y./\z.yap(xap(f, y), z)) => x rec(s(x), y, /\z./\u.yap(xap(f, z), u)) => yap(xap(f, x), rec(x, y, /\v./\w.yap(xap(f, v), w))) 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 rec : [nat * a * nat -> a -> a] --> a s : [nat] --> nat yap : [a -> a * a] --> a Rules: rec(0, X, /\x./\y.yap(F(x), y)) => X rec(s(X), Y, /\x./\y.yap(F(x), y)) => yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) 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(0, X, /\x./\y.yap(F(x), y)) >? X rec(s(X), Y, /\x./\y.yap(F(x), y)) >? yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) 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(0, X, /\x./\y.yap(F(x), y)) > X because [2], by definition 2] rec*(0, X, /\x./\y.yap(F(x), y)) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because [5], by (Star) 5] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because rec > yap, [6] and [13], by (Copy) 6] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= F(X) because [7], by (Select) 7] /\x.yap(F(rec*(s(X), Y, /\y./\z.yap(F(y), z))), x) >= F(X) because [8], by (Eta)[Kop13:2] 8] F(rec*(s(X), Y, /\x./\y.yap(F(x), y))) >= F(X) because [9], by (Meta) 9] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= X because [10], by (Select) 10] s(X) >= X because [11], by (Star) 11] s*(X) >= X because [12], by (Select) 12] X >= X by (Meta) 13] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= rec(X, Y, /\x./\y.yap(F(x), y)) because rec in Mul, [14], [16] and [17], by (Stat) 14] s(X) > X because [15], by definition 15] s*(X) >= X because [12], by (Select) 16] Y >= Y by (Meta) 17] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [18], by (Abs) 18] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [19], by (Abs) 19] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [20] and [22], by (Fun) 20] F(y) >= F(y) because [21], by (Meta) 21] y >= y by (Var) 22] x >= x by (Var) 23] yap(F, X) >= @_{o -> o}(F, X) because [24], by (Star) 24] yap*(F, X) >= @_{o -> o}(F, X) because yap > @_{o -> o}, [25] and [27], by (Copy) 25] yap*(F, X) >= F because [26], by (Select) 26] F >= F by (Meta) 27] yap*(F, X) >= X because [28], by (Select) 28] X >= X by (Meta) We can thus remove the following rules: rec(0, X, /\x./\y.yap(F(x), y)) => 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(s(X), Y, /\x./\y.yap(F(x), y)) >? yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o}, rec, s, yap}, and the following precedence: rec > s > yap > @_{o -> o} With these choices, we have: 1] rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because [2], by (Star) 2] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because rec > yap, [3] and [10], by (Copy) 3] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= F(X) because [4], by (Select) popout

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all output return to Higher Order Rewriting Union Beta