details

property	value
status	complete
benchmark	Applicative_05__Ex6Folding.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.66192 seconds
cpu usage	0.661765
user time	0.598613
system time	0.063152
max virtual memory	132724.0
max residence set size	18532.0

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 Rules: fold(f, nil, x) => x fold(f, cons(x, y), z) => fold(f, y, f z x) sum(x) => fold(add, x, 0) fold(mul, x, 1) => prod(x) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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(F, nil, X) >? X fold(F, cons(X, Y), Z) >? fold(F, Y, F Z X) sum(X) >? fold(add, X, 0) fold(mul, X, 1) >? prod(X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[add]] = _|_ [[fold(x_1, x_2, x_3)]] = fold(x_2, x_1, x_3) We choose Lex = {fold, prod} and Mul = {1, @_{o -> o -> o}, @_{o -> o}, cons, mul, nil, sum}, and the following precedence: 1 > cons > mul > nil > sum > fold = prod > @_{o -> o -> o} > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: fold(F, nil, X) >= X fold(F, cons(X, Y), Z) >= fold(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X)) sum(X) >= fold(_|_, X, _|_) fold(mul, X, 1) > prod(X) With these choices, we have: 1] fold(F, nil, X) >= X because [2], by (Star) 2] fold*(F, nil, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] fold(F, cons(X, Y), Z) >= fold(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X)) because [5], by (Star) 5] fold*(F, cons(X, Y), Z) >= fold(F, Y, @_{o -> o}(@_{o -> o -> o}(F, Z), X)) because [6], [9], [11] and [13], by (Stat) 6] cons(X, Y) > Y because [7], by definition 7] cons*(X, Y) >= Y because [8], by (Select) 8] Y >= Y by (Meta) 9] fold*(F, cons(X, Y), Z) >= F because [10], by (Select) 10] F >= F by (Meta) 11] fold*(F, cons(X, Y), Z) >= Y because [12], by (Select) 12] cons(X, Y) >= Y because [7], by (Star) 13] fold*(F, cons(X, Y), Z) >= @_{o -> o}(@_{o -> o -> o}(F, Z), X) because fold > @_{o -> o}, [14] and [17], by (Copy) 14] fold*(F, cons(X, Y), Z) >= @_{o -> o -> o}(F, Z) because fold > @_{o -> o -> o}, [9] and [15], by (Copy) 15] fold*(F, cons(X, Y), Z) >= Z because [16], by (Select) 16] Z >= Z by (Meta) 17] fold*(F, cons(X, Y), Z) >= X because [18], by (Select) 18] cons(X, Y) >= X because [19], by (Star) 19] cons*(X, Y) >= X because [20], by (Select) 20] X >= X by (Meta) 21] sum(X) >= fold(_|_, X, _|_) because [22], by (Star) 22] sum*(X) >= fold(_|_, X, _|_) because sum > fold, [23], [24] and [26], by (Copy) 23] sum*(X) >= _|_ by (Bot) 24] sum*(X) >= X because [25], by (Select) 25] X >= X by (Meta) 26] sum*(X) >= _|_ by (Bot) 27] fold(mul, X, 1) > prod(X) because [28], by definition 28] fold*(mul, X, 1) >= prod(X) because fold = prod, [29] and [30], by (Stat) 29] X >= X by (Meta) 30] fold*(mul, X, 1) >= X because [29], by (Select) We can thus remove the following rules: fold(mul, X, 1) => prod(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(F, nil, X) >? X fold(F, cons(X, Y), Z) >? fold(F, Y, F Z X) sum(X) >? fold(add, X, 0) We use a recursive path ordering as defined in [Kop12, Chapter 5]. popout

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job log

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actions

all output return to Higher Order Rewriting Union Beta