Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270299

details

property	value
status	complete
benchmark	length.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n174.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.167411 seconds
cpu usage	0.167687
user time	0.138234
system time	0.029453
max virtual memory	113176.0
max residence set size	5760.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.16	YES
0.00/0.16	We consider the system theBenchmark.
0.00/0.16	
0.00/0.16	  Alphabet:
0.00/0.16	
0.00/0.16	    0 : [] --> nat 
0.00/0.16	    cons : [nat * list] --> list 
0.00/0.16	    foldr : [nat -> nat -> nat * nat * list] --> nat 
0.00/0.16	    length : [list] --> nat 
0.00/0.16	    nil : [] --> list 
0.00/0.16	    s : [nat] --> nat 
0.00/0.16	
0.00/0.16	  Rules:
0.00/0.16	
0.00/0.16	    foldr(f, x, nil) => x 
0.00/0.16	    foldr(f, x, cons(y, z)) => f y foldr(f, x, z) 
0.00/0.16	    length(x) => foldr(/\y./\z.s(z), 0, x) 
0.00/0.16	
0.00/0.16	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.16	
0.00/0.16	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.16	
0.00/0.16	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.16	
0.00/0.16	  foldr(F, X, nil) >? X 
0.00/0.16	  foldr(F, X, cons(Y, Z)) >? F Y foldr(F, X, Z) 
0.00/0.16	  length(X) >? foldr(/\x./\y.s(y), 0, X) 
0.00/0.16	
0.00/0.16	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.16	
0.00/0.16	Argument functions:
0.00/0.16	
0.00/0.16	  [[0]] = _|_ 
0.00/0.16	  [[s(x_1)]] = x_1 
0.00/0.16	
0.00/0.16	We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, cons, foldr, length, nil}, and the following precedence: length > foldr > @_{o -> o} > @_{o -> o -> o} > cons > nil
0.00/0.16	
0.00/0.16	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.16	
0.00/0.16	  foldr(F, X, nil) >= X 
0.00/0.16	  foldr(F, X, cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z)) 
0.00/0.16	  length(X) >= foldr(/\x./\y.y, _|_, X) 
0.00/0.16	
0.00/0.16	With these choices, we have:
0.00/0.16	
0.00/0.16	  1] foldr(F, X, nil) >= X  because [2], by (Star) 
0.00/0.16	  2] foldr*(F, X, nil) >= X  because [3], by (Select) 
0.00/0.16	  3] X >= X  by (Meta) 
0.00/0.16	
0.00/0.16	  4] foldr(F, X, cons(Y, Z)) > @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because [5], by definition 
0.00/0.16	  5] foldr*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, Y), foldr(F, X, Z))  because foldr > @_{o -> o}, [6] and [13], by (Copy) 
0.00/0.16	  6] foldr*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, Y)  because foldr > @_{o -> o -> o}, [7] and [9], by (Copy) 
0.00/0.16	  7] foldr*(F, X, cons(Y, Z)) >= F  because [8], by (Select) 
0.00/0.16	  8] F >= F  by (Meta) 
0.00/0.16	  9] foldr*(F, X, cons(Y, Z)) >= Y  because [10], by (Select) 
0.00/0.16	  10] cons(Y, Z) >= Y  because [11], by (Star) 
0.00/0.16	  11] cons*(Y, Z) >= Y  because [12], by (Select) 
0.00/0.16	  12] Y >= Y  by (Meta) 
0.00/0.16	  13] foldr*(F, X, cons(Y, Z)) >= foldr(F, X, Z)  because foldr in Mul, [14], [15] and [16], by (Stat) 
0.00/0.16	  14] F >= F  by (Meta) 
0.00/0.16	  15] X >= X  by (Meta) 
0.00/0.16	  16] cons(Y, Z) > Z  because [17], by definition 
0.00/0.16	  17] cons*(Y, Z) >= Z  because [18], by (Select) 
0.00/0.16	  18] Z >= Z  by (Meta) 
0.00/0.16	
0.00/0.16	  19] length(X) >= foldr(/\x./\y.y, _|_, X)  because [20], by (Star) 
0.00/0.16	  20] length*(X) >= foldr(/\x./\y.y, _|_, X)  because length > foldr, [21], [25] and [26], by (Copy) 
0.00/0.16	  21] length*(X) >= /\y./\z.z  because [22], by (F-Abs) 
0.00/0.16	  22] length*(X, x) >= /\z.z  because [23], by (F-Abs) 
0.00/0.16	  23] length*(X, x, y) >= y  because [24], by (Select) 
0.00/0.16	  24] y >= y  by (Var) 
0.00/0.16	  25] length*(X) >= _|_  by (Bot) 
0.00/0.16	  26] length*(X) >= X  because [27], by (Select) 
0.00/0.16	  27] X >= X  by (Meta) 
0.00/0.16	
0.00/0.16	We can thus remove the following rules:
0.00/0.16	
0.00/0.16	  foldr(F, X, cons(Y, Z)) => F Y foldr(F, X, Z) 
0.00/0.16	
0.00/0.16	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.16	
0.00/0.16	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.16	
0.00/0.16	  foldr(F, X, nil) >? X 
0.00/0.16	  length(X) >? foldr(/\x./\y.s(y), 0, X) 
0.00/0.16	
0.00/0.16	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.16	
0.00/0.16	The following interpretation satisfies the requirements:
0.00/0.16	
0.00/0.16	  0 = 0 
0.00/0.16	  foldr = \G0y1y2.y1 + y2 + G0(0,0) 
0.00/0.16	  length = \y0.3 + 3y0 
0.00/0.16	  nil = 3 
0.00/0.16	  s = \y0.y0 
0.00/0.16	
0.00/0.16	Using this interpretation, the requirements translate to:
0.00/0.16	
0.00/0.16	  [[foldr(_F0, _x1, nil)]] = 3 + x1 + F0(0,0) > x1 = [[_x1]] 
0.00/0.16	  [[length(_x0)]] = 3 + 3x0 > x0 = [[foldr(/\x./\y.s(y), 0, _x0)]]

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actions all output return to Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10