details

property	value
status	complete
benchmark	h03.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)	4.59028 seconds
cpu usage	4.58988
user time	4.49381
system time	0.096069
max virtual memory	168364.0
max residence set size	51772.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> c 1 : [] --> c add : [] --> c -> a -> c cons : [a * b] --> b fold : [c -> a -> c * b * c] --> c mul : [] --> c -> a -> c nil : [] --> b prod : [b] --> c sum : [b] --> c xap : [c -> a -> c * c] --> a -> c yap : [a -> c * a] --> c Rules: fold(/\x./\y.yap(xap(f, x), y), nil, z) => z fold(/\x./\y.yap(xap(f, x), y), cons(z, u), v) => fold(/\w./\x'.yap(xap(f, w), x'), u, yap(xap(f, v), z)) sum(x) => fold(/\y./\z.yap(xap(add, y), z), x, 0) fold(mul, x, 1) => prod(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 : [] --> c 1 : [] --> c add : [c] --> a -> c cons : [a * b] --> b fold : [c -> a -> c * b * c] --> c mul : [] --> c -> a -> c nil : [] --> b prod : [b] --> c sum : [b] --> c yap : [a -> c * a] --> c Rules: fold(/\x./\y.yap(F(x), y), nil, X) => X fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) => fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) sum(X) => fold(/\x./\y.yap(add(x), y), X, 0) fold(mul, X, 1) => prod(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]): fold(/\x./\y.yap(F(x), y), nil, X) >? X fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) >? fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) sum(X) >? fold(/\x./\y.yap(add(x), y), X, 0) fold(mul, X, 1) >? prod(X) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[fold(x_1, x_2, x_3)]] = fold(x_2, x_3, x_1) [[prod(x_1)]] = x_1 We choose Lex = {fold} and Mul = {1, @_{o -> o}, add, cons, mul, nil, sum, yap}, and the following precedence: 1 > mul > nil > cons > sum > add > fold > yap > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: fold(/\x./\y.yap(F(x), y), nil, X) >= X fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) > fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) sum(X) >= fold(/\x./\y.yap(add(x), y), X, _|_) fold(mul, X, 1) > X yap(F, X) >= @_{o -> o}(F, X) With these choices, we have: 1] fold(/\x./\y.yap(F(x), y), nil, X) >= X because [2], by (Star) 2] fold*(/\x./\y.yap(F(x), y), nil, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) > fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) because [5], by definition 5] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) because [6], by (Select) 6] yap(F(fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z)), fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) because [7], by (Star) 7] yap*(F(fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z)), fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) because [8], by (Select) 8] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) because [9], [12], [19] and [21], by (Stat) 9] cons(X, Y) > Y because [10], by definition 10] cons*(X, Y) >= Y because [11], by (Select) 11] Y >= Y by (Meta) 12] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= /\x./\y.yap(F(x), y) because [13], by (Select) 13] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [14], by (Abs) 14] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [15], by (Abs) 15] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [16] and [18], by (Fun) 16] F(y) >= F(y) because [17], by (Meta) 17] y >= y by (Var) 18] x >= x by (Var) popout

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all output return to Higher Order Rewriting Union Beta