details

property	value
status	complete
benchmark	h51.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.241275 seconds
cpu usage	0.24156
user time	0.216636
system time	0.024924
max virtual memory	113188.0
max residence set size	10260.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: cons : [] --> a -> alist -> alist foldl : [] --> (a -> a -> a) -> a -> alist -> a nil : [] --> alist Rules: foldl (/\x./\y.f x y) z nil => z foldl (/\x./\y.f x y) z (cons u v) => foldl (/\w./\x'.f w x') (f z u) v Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: Alphabet: cons : [a * alist] --> alist foldl : [a -> a -> a * a * alist] --> a nil : [] --> alist ~AP1 : [a -> a -> a * a] --> a -> a Rules: foldl(/\x./\y.~AP1(F, x) y, X, nil) => X foldl(/\x./\y.~AP1(F, x) y, X, cons(Y, Z)) => foldl(/\z./\u.~AP1(F, z) u, ~AP1(F, X) Y, Z) ~AP1(F, X) => F X Symbol ~AP1 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: cons : [a * alist] --> alist foldl : [a -> a -> a * a * alist] --> a nil : [] --> alist Rules: foldl(/\x./\y.X(x, y), Y, nil) => Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) => foldl(/\z./\u.X(z, u), X(Y, Z), U) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): foldl(/\x./\y.X(x, y), Y, nil) >? Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldl(/\z./\u.X(z, u), X(Y, Z), U) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_2, x_1) We choose Lex = {foldl} and Mul = {cons, nil}, and the following precedence: cons > foldl > nil Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.X(x, y), Y, nil) > Y foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) With these choices, we have: 1] foldl(/\x./\y.X(x, y), Y, nil) > Y because [2], by definition 2] foldl*(/\x./\y.X(x, y), Y, nil) >= Y because [3], by (Select) 3] Y >= Y by (Meta) 4] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [5], by (Star) 5] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) because [6], [9], [16] and [24], by (Stat) 6] cons(Z, U) > U because [7], by definition 7] cons*(Z, U) >= U because [8], by (Select) 8] U >= U by (Meta) 9] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y) because [10], by (F-Abs) 10] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= /\x.X(z, x) because [11], by (Select) 11] /\x.X(foldl*(/\y./\v.X(y, v), Y, cons(Z, U), z), x) >= /\x.X(z, x) because [12], by (Abs) 12] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z), u) >= X(z, u) because [13] and [15], by (Meta) 13] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= z because [14], by (Select) 14] z >= z by (Var) 15] u >= u by (Var) 16] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Y, Z) because [17], by (Select) 17] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z) because [18] and [20], by (Meta) 18] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Y because [19], by (Select) 19] Y >= Y by (Meta) 20] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z because [21], by (Select) 21] cons(Z, U) >= Z because [22], by (Star) 22] cons*(Z, U) >= Z because [23], by (Select) 23] Z >= Z by (Meta) 24] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= U because [25], by (Select) 25] cons(Z, U) >= U because [7], by (Star) We can thus remove the following rules: foldl(/\x./\y.X(x, y), Y, nil) => Y We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): popout

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all output return to Higher Order Rewriting Union Beta