Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270217

details

property	value
status	complete
benchmark	kop12thesis_ex2.11.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n155.star.cs.uiowa.edu
space	Kop_13

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.344905 seconds
cpu usage	0.34539
user time	0.298102
system time	0.047288
max virtual memory	113176.0
max residence set size	11628.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.32	YES
0.00/0.34	We consider the system theBenchmark.
0.00/0.34	
0.00/0.34	  Alphabet:
0.00/0.34	
0.00/0.34	    cons : [nat * list] --> list 
0.00/0.34	    emap : [nat -> nat * list] --> list 
0.00/0.34	    nil : [] --> list 
0.00/0.34	    twice : [nat -> nat] --> nat -> nat 
0.00/0.34	
0.00/0.34	  Rules:
0.00/0.34	
0.00/0.34	    emap(f, nil) => nil 
0.00/0.34	    emap(f, cons(x, y)) => cons(f x, emap(/\z.twice(f) z, y)) 
0.00/0.34	    twice(f) x => f (f x) 
0.00/0.34	
0.00/0.34	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.34	
0.00/0.34	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.34	
0.00/0.34	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.34	
0.00/0.34	  emap(F, nil) >? nil 
0.00/0.34	  emap(F, cons(X, Y)) >? cons(F X, emap(/\x.twice(F, x), Y)) 
0.00/0.34	  twice(F, X) >? F (F X) 
0.00/0.34	
0.00/0.34	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.34	
0.00/0.34	Argument functions:
0.00/0.34	
0.00/0.34	  [[emap(x_1, x_2)]] = emap(x_2, x_1) 
0.00/0.34	  [[nil]] = _|_ 
0.00/0.34	
0.00/0.34	We choose Lex = {emap} and Mul = {@_{o -> o}, cons, twice}, and the following precedence: emap > cons > twice > @_{o -> o}
0.00/0.34	
0.00/0.34	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.34	
0.00/0.34	  emap(F, _|_) > _|_ 
0.00/0.34	  emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) 
0.00/0.34	  twice(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X)) 
0.00/0.34	
0.00/0.34	With these choices, we have:
0.00/0.34	
0.00/0.34	  1] emap(F, _|_) > _|_  because [2], by definition 
0.00/0.34	  2] emap*(F, _|_) >= _|_  by (Bot) 
0.00/0.34	
0.00/0.34	  3] emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y))  because [4], by (Star) 
0.00/0.34	  4] emap*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y))  because emap > cons, [5] and [12], by (Copy) 
0.00/0.34	  5] emap*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because emap > @_{o -> o}, [6] and [8], by (Copy) 
0.00/0.34	  6] emap*(F, cons(X, Y)) >= F  because [7], by (Select) 
0.00/0.34	  7] F >= F  by (Meta) 
0.00/0.34	  8] emap*(F, cons(X, Y)) >= X  because [9], by (Select) 
0.00/0.34	  9] cons(X, Y) >= X  because [10], by (Star) 
0.00/0.34	  10] cons*(X, Y) >= X  because [11], by (Select) 
0.00/0.34	  11] X >= X  by (Meta) 
0.00/0.34	  12] emap*(F, cons(X, Y)) >= emap(/\x.twice(F, x), Y)  because [13], [16] and [21], by (Stat) 
0.00/0.34	  13] cons(X, Y) > Y  because [14], by definition 
0.00/0.34	  14] cons*(X, Y) >= Y  because [15], by (Select) 
0.00/0.34	  15] Y >= Y  by (Meta) 
0.00/0.34	  16] emap*(F, cons(X, Y)) >= /\y.twice(F, y)  because [17], by (F-Abs) 
0.00/0.34	  17] emap*(F, cons(X, Y), x) >= twice(F, x)  because emap > twice, [18] and [19], by (Copy) 
0.00/0.34	  18] emap*(F, cons(X, Y), x) >= F  because [7], by (Select) 
0.00/0.34	  19] emap*(F, cons(X, Y), x) >= x  because [20], by (Select) 
0.00/0.34	  20] x >= x  by (Var) 
0.00/0.34	  21] emap*(F, cons(X, Y)) >= Y  because [22], by (Select) 
0.00/0.34	  22] cons(X, Y) >= Y  because [14], by (Star) 
0.00/0.34	
0.00/0.34	  23] twice(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X))  because [24], by (Star) 
0.00/0.34	  24] twice*(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X))  because twice > @_{o -> o}, [25] and [27], by (Copy) 
0.00/0.34	  25] twice*(F, X) >= F  because [26], by (Select) 
0.00/0.34	  26] F >= F  by (Meta) 
0.00/0.34	  27] twice*(F, X) >= @_{o -> o}(F, X)  because twice > @_{o -> o}, [25] and [28], by (Copy) 
0.00/0.34	  28] twice*(F, X) >= X  because [29], by (Select) 
0.00/0.34	  29] X >= X  by (Meta) 
0.00/0.34	
0.00/0.34	We can thus remove the following rules:
0.00/0.34	
0.00/0.34	  emap(F, nil) => nil 
0.00/0.34	
0.00/0.34	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.34	
0.00/0.34	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.34	
0.00/0.34	  emap(F, cons(X, Y)) >? cons(F X, emap(/\x.twice(F, x), Y)) 
0.00/0.34	  twice(F, X) >? F (F X) 
0.00/0.34	
0.00/0.34	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.34	
0.00/0.34	Argument functions:
0.00/0.34	
0.00/0.34	  [[emap(x_1, x_2)]] = emap(x_2, x_1) 
0.00/0.34	
0.00/0.34	We choose Lex = {emap} and Mul = {@_{o -> o}, cons, twice}, and the following precedence: emap > cons > twice > @_{o -> o}
0.00/0.34	
0.00/0.34	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.34	
0.00/0.34	  emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) 
0.00/0.34	  twice(F, X) > @_{o -> o}(F, @_{o -> o}(F, X)) 
0.00/0.34	
0.00/0.34	With these choices, we have:

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