details

property	value
status	complete
benchmark	h01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	31.7136 seconds
cpu usage	31.0383
user time	30.419
system time	0.619332
max virtual memory	748800.0
max residence set size	404408.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> c add : [] --> a -> c -> c cons : [a * b] --> b fold : [a -> c -> c * c] --> b -> c mul : [] --> a -> c -> c nil : [] --> b plus : [c * c] --> c prod : [] --> b -> c s : [c] --> c sum : [] --> b -> c times : [c * c] --> c xap : [a -> c -> c * a] --> c -> c yap : [c -> c * c] --> c Rules: fold(/\x./\y.yap(xap(f, x), y), z) nil => z fold(/\x./\y.yap(xap(f, x), y), z) cons(u, v) => yap(xap(f, u), fold(/\w./\x'.yap(xap(f, w), x'), z) v) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) times(0, x) => 0 times(s(x), y) => plus(times(x, y), y) sum => fold(/\x./\y.yap(xap(add, x), y), 0) prod => fold(/\x./\y.yap(xap(mul, x), y), s(0)) 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 add : [a] --> c -> c cons : [a * b] --> b fold : [a -> c -> c * c] --> b -> c mul : [a] --> c -> c nil : [] --> b plus : [c * c] --> c prod : [] --> b -> c s : [c] --> c sum : [] --> b -> c times : [c * c] --> c yap : [c -> c * c] --> c Rules: fold(/\x./\y.yap(F(x), y), X) nil => X fold(/\x./\y.yap(F(x), y), X) cons(Y, Z) => yap(F(Y), fold(/\z./\u.yap(F(z), u), X) Z) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(times(X, Y), Y) sum => fold(/\x./\y.yap(add(x), y), 0) prod => fold(/\x./\y.yap(mul(x), y), s(0)) 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), X) nil >? X fold(/\x./\y.yap(F(x), y), X) cons(Y, Z) >? yap(F(Y), fold(/\z./\u.yap(F(z), u), X) Z) plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) times(0, X) >? 0 times(s(X), Y) >? plus(times(X, Y), Y) sum >? fold(/\x./\y.yap(add(x), y), 0) prod >? fold(/\x./\y.yap(mul(x), y), s(0)) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, add, cons, fold, mul, nil, plus, prod, s, sum, times, yap}, and the following precedence: cons > prod > mul > nil > sum > add > fold > times > plus > s > yap > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(fold(/\x./\y.yap(F(x), y), X), nil) >= X @_{o -> o}(fold(/\x./\y.yap(F(x), y), X), cons(Y, Z)) >= yap(F(Y), @_{o -> o}(fold(/\x./\y.yap(F(x), y), X), Z)) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(_|_, X) >= _|_ times(s(X), Y) > plus(times(X, Y), Y) sum >= fold(/\x./\y.yap(add(x), y), _|_) prod >= fold(/\x./\y.yap(mul(x), y), s(_|_)) yap(F, X) >= @_{o -> o}(F, X) With these choices, we have: 1] @_{o -> o}(fold(/\x./\y.yap(F(x), y), X), nil) >= X because [2], by (Star) popout

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actions

all output return to Higher Order Rewriting Union Beta