Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270531

details

property	value
status	complete
benchmark	AotoYamada_05__015.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n086.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.339799 seconds
cpu usage	0.340383
user time	0.294168
system time	0.046215
max virtual memory	113176.0
max residence set size	9816.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.32	YES
0.00/0.33	We consider the system theBenchmark.
0.00/0.33	
0.00/0.33	  Alphabet:
0.00/0.33	
0.00/0.33	    cons : [a * c] --> c 
0.00/0.33	    false : [] --> b 
0.00/0.33	    filter : [a -> b * c] --> c 
0.00/0.33	    if : [b * c * c] --> c 
0.00/0.33	    nil : [] --> c 
0.00/0.33	    true : [] --> b 
0.00/0.33	
0.00/0.33	  Rules:
0.00/0.33	
0.00/0.33	    if(true, x, y) => x 
0.00/0.33	    if(false, x, y) => y 
0.00/0.33	    filter(f, nil) => nil 
0.00/0.33	    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(f, y)) 
0.00/0.33	
0.00/0.33	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.33	
0.00/0.33	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.33	
0.00/0.33	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.33	
0.00/0.33	  if(true, X, Y) >? X 
0.00/0.33	  if(false, X, Y) >? Y 
0.00/0.33	  filter(F, nil) >? nil 
0.00/0.33	  filter(F, cons(X, Y)) >? if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
0.00/0.33	
0.00/0.33	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.33	
0.00/0.33	Argument functions:
0.00/0.33	
0.00/0.33	  [[nil]] = _|_ 
0.00/0.33	
0.00/0.33	We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, if, true}, and the following precedence: false > filter > @_{o -> o} > cons > if > true
0.00/0.33	
0.00/0.33	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.33	
0.00/0.33	  if(true, X, Y) >= X 
0.00/0.33	  if(false, X, Y) >= Y 
0.00/0.33	  filter(F, _|_) > _|_ 
0.00/0.33	  filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y)) 
0.00/0.33	
0.00/0.33	With these choices, we have:
0.00/0.33	
0.00/0.33	  1] if(true, X, Y) >= X  because [2], by (Star) 
0.00/0.33	  2] if*(true, X, Y) >= X  because [3], by (Select) 
0.00/0.33	  3] X >= X  by (Meta) 
0.00/0.33	
0.00/0.33	  4] if(false, X, Y) >= Y  because [5], by (Star) 
0.00/0.33	  5] if*(false, X, Y) >= Y  because [6], by (Select) 
0.00/0.33	  6] Y >= Y  by (Meta) 
0.00/0.33	
0.00/0.33	  7] filter(F, _|_) > _|_  because [8], by definition 
0.00/0.33	  8] filter*(F, _|_) >= _|_  by (Bot) 
0.00/0.33	
0.00/0.33	  9] filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because [10], by (Star) 
0.00/0.33	  10] filter*(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because filter > if, [11], [18] and [19], by (Copy) 
0.00/0.33	  11] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [12] and [14], by (Copy) 
0.00/0.33	  12] filter*(F, cons(X, Y)) >= F  because [13], by (Select) 
0.00/0.33	  13] F >= F  by (Meta) 
0.00/0.33	  14] filter*(F, cons(X, Y)) >= X  because [15], by (Select) 
0.00/0.33	  15] cons(X, Y) >= X  because [16], by (Star) 
0.00/0.33	  16] cons*(X, Y) >= X  because [17], by (Select) 
0.00/0.33	  17] X >= X  by (Meta) 
0.00/0.33	  18] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y))  because filter > cons, [14] and [19], by (Copy) 
0.00/0.33	  19] filter*(F, cons(X, Y)) >= filter(F, Y)  because filter in Mul, [20] and [21], by (Stat) 
0.00/0.33	  20] F >= F  by (Meta) 
0.00/0.33	  21] cons(X, Y) > Y  because [22], by definition 
0.00/0.33	  22] cons*(X, Y) >= Y  because [23], by (Select) 
0.00/0.33	  23] Y >= Y  by (Meta) 
0.00/0.33	
0.00/0.33	We can thus remove the following rules:
0.00/0.33	
0.00/0.33	  filter(F, nil) => nil 
0.00/0.33	
0.00/0.33	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.33	
0.00/0.33	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.33	
0.00/0.33	  if(true, X, Y) >? X 
0.00/0.33	  if(false, X, Y) >? Y 
0.00/0.33	  filter(F, cons(X, Y)) >? if(F X, cons(X, filter(F, Y)), filter(F, Y)) 
0.00/0.33	
0.00/0.33	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.33	
0.00/0.33	We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, if, true}, and the following precedence: false > filter > @_{o -> o} > if > cons > true
0.00/0.33	
0.00/0.33	With these choices, we have:
0.00/0.33	
0.00/0.33	  1] if(true, X, Y) >= X  because [2], by (Star) 
0.00/0.33	  2] if*(true, X, Y) >= X  because [3], by (Select) 
0.00/0.33	  3] X >= X  by (Meta) 
0.00/0.33	
0.00/0.33	  4] if(false, X, Y) > Y  because [5], by definition 
0.00/0.33	  5] if*(false, X, Y) >= Y  because [6], by (Select) 
0.00/0.33	  6] Y >= Y  by (Meta) 
0.00/0.33

popout

output may be truncated. 'popout' for the full output.

job log

popout

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