details

property	value
status	complete
benchmark	h54.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	9.30678 seconds
cpu usage	9.30706
user time	9.14981
system time	0.15725
max virtual memory	213800.0
max residence set size	99400.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat cons : [nat * list] --> list foldl : [nat -> nat -> nat * nat * list] --> nat nil : [] --> list plusc : [] --> nat -> nat -> nat s : [nat] --> nat sum : [list] --> nat xap : [nat -> nat -> nat * nat] --> nat -> nat yap : [nat -> nat * nat] --> nat Rules: foldl(/\x./\y.yap(xap(f, x), y), z, nil) => z foldl(/\x./\y.yap(xap(f, x), y), z, cons(u, v)) => foldl(/\w./\x'.yap(xap(f, w), x'), yap(xap(f, z), u), v) plusc x 0 => x plusc x s(y) => s(plusc x y) sum(x) => foldl(/\y./\z.yap(xap(plusc, y), z), 0, 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 : [] --> nat cons : [nat * list] --> list foldl : [nat -> nat -> nat * nat * list] --> nat nil : [] --> list plusc : [nat] --> nat -> nat s : [nat] --> nat sum : [list] --> nat yap : [nat -> nat * nat] --> nat Rules: foldl(/\x./\y.yap(F(x), y), X, nil) => X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) => foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) plusc(X) 0 => X plusc(X) s(Y) => s(plusc(X) Y) sum(X) => foldl(/\x./\y.yap(plusc(x), y), 0, 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]): foldl(/\x./\y.yap(F(x), y), X, nil) >? X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >? foldl(/\z./\u.yap(F(z), u), yap(F(X), Y), Z) plusc(X) 0 >? X plusc(X) s(Y) >? s(plusc(X) Y) sum(X) >? foldl(/\x./\y.yap(plusc(x), y), 0, X) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) [[s(x_1)]] = x_1 We choose Lex = {foldl} and Mul = {@_{o -> o}, cons, nil, plusc, sum, yap}, and the following precedence: nil > sum > plusc > cons > yap > @_{o -> o} > foldl Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: foldl(/\x./\y.yap(F(x), y), X, nil) >= X foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) @_{o -> o}(plusc(X), _|_) >= X @_{o -> o}(plusc(X), Y) >= @_{o -> o}(plusc(X), Y) sum(X) >= foldl(/\x./\y.yap(plusc(x), y), _|_, X) yap(F, X) > @_{o -> o}(F, X) With these choices, we have: 1] foldl(/\x./\y.yap(F(x), y), X, nil) >= X because [2], by (Star) 2] foldl*(/\x./\y.yap(F(x), y), X, nil) >= X because [3], by (Select) 3] X >= X by (Meta) 4] foldl(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [5], by (Star) 5] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= foldl(/\x./\y.yap(F(x), y), yap(F(X), Y), Z) because [6], [9], [16] and [25], by (Stat) 6] cons(Y, Z) > Z because [7], by definition 7] cons*(Y, Z) >= Z because [8], by (Select) 8] Z >= Z by (Meta) 9] foldl*(/\x./\y.yap(F(x), y), X, cons(Y, Z)) >= /\x./\y.yap(F(x), y) because [10], by (Select) 10] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [11], by (Abs) 11] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [12], by (Abs) 12] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [13] and [15], by (Fun) 13] F(y) >= F(y) because [14], by (Meta) 14] y >= y by (Var) 15] x >= x by (Var) 16] foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z)) >= yap(F(X), Y) because [17], by (Select) 17] yap(F(foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z))), foldl*(/\v./\w.yap(F(v), w), X, cons(Y, Z))) >= yap(F(X), Y) because yap in Mul, [18] and [21], by (Fun) 18] F(foldl*(/\z./\u.yap(F(z), u), X, cons(Y, Z))) >= F(X) because [19], by (Meta) popout

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actions

all output return to Higher Order Rewriting Union Beta