Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270663

details

property	value
status	complete
benchmark	Applicative_first_order_05__#3.27.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n059.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.185212 seconds
cpu usage	0.185349
user time	0.171284
system time	0.014065
max virtual memory	113176.0
max residence set size	4388.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.17	YES
0.00/0.18	We consider the system theBenchmark.
0.00/0.18	
0.00/0.18	  Alphabet:
0.00/0.18	
0.00/0.18	    cons : [c * d] --> d 
0.00/0.18	    f : [a] --> a 
0.00/0.18	    false : [] --> b 
0.00/0.18	    filter : [c -> b * d] --> d 
0.00/0.18	    filter2 : [b * c -> b * c * d] --> d 
0.00/0.18	    g : [a] --> a 
0.00/0.18	    map : [c -> c * d] --> d 
0.00/0.18	    nil : [] --> d 
0.00/0.18	    true : [] --> b 
0.00/0.18	
0.00/0.18	  Rules:
0.00/0.18	
0.00/0.18	    f(f(x)) => g(f(x)) 
0.00/0.18	    g(g(x)) => f(x) 
0.00/0.18	    map(h, nil) => nil 
0.00/0.18	    map(h, cons(x, y)) => cons(h x, map(h, y)) 
0.00/0.18	    filter(h, nil) => nil 
0.00/0.18	    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
0.00/0.18	    filter2(true, h, x, y) => cons(x, filter(h, y)) 
0.00/0.18	    filter2(false, h, x, y) => filter(h, y) 
0.00/0.18	
0.00/0.18	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.18	
0.00/0.18	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.18	
0.00/0.18	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.18	
0.00/0.18	  f(f(X)) >? g(f(X)) 
0.00/0.18	  g(g(X)) >? f(X) 
0.00/0.18	  map(F, nil) >? nil 
0.00/0.18	  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
0.00/0.18	  filter(F, nil) >? nil 
0.00/0.18	  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
0.00/0.18	  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
0.00/0.18	  filter2(false, F, X, Y) >? filter(F, Y) 
0.00/0.18	
0.00/0.18	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.18	
0.00/0.18	The following interpretation satisfies the requirements:
0.00/0.18	
0.00/0.18	  cons = \y0y1.1 + y0 + y1 
0.00/0.18	  f = \y0.2y0 
0.00/0.18	  false = 3 
0.00/0.18	  filter = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
0.00/0.18	  filter2 = \y0G1y2y3.y2 + 2y0 + 3y3 + G1(0) + 2y3G1(y3) 
0.00/0.18	  g = \y0.2y0 
0.00/0.18	  map = \G0y1.2 + 3y1 + G0(y1) + y1G0(y1) 
0.00/0.18	  nil = 0 
0.00/0.18	  true = 3 
0.00/0.18	
0.00/0.18	Using this interpretation, the requirements translate to:
0.00/0.18	
0.00/0.18	  [[f(f(_x0))]] = 4x0 >= 4x0 = [[g(f(_x0))]] 
0.00/0.18	  [[g(g(_x0))]] = 4x0 >= 2x0 = [[f(_x0)]] 
0.00/0.18	  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
0.00/0.18	  [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + F0(x2) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.18	  [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
0.00/0.18	  [[filter(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 3x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
0.00/0.18	  [[filter2(true, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 1 + x1 + 3x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
0.00/0.18	  [[filter2(false, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 3x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 
0.00/0.18	
0.00/0.18	We can thus remove the following rules:
0.00/0.18	
0.00/0.18	  map(F, nil) => nil 
0.00/0.18	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.18	  filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
0.00/0.18	  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
0.00/0.18	  filter2(false, F, X, Y) => filter(F, Y) 
0.00/0.18	
0.00/0.18	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.18	
0.00/0.18	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.18	
0.00/0.18	  f(f(X)) >? g(f(X)) 
0.00/0.18	  g(g(X)) >? f(X) 
0.00/0.18	  filter(F, nil) >? nil 
0.00/0.18	
0.00/0.18	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.18	
0.00/0.18	The following interpretation satisfies the requirements:
0.00/0.18	
0.00/0.18	  f = \y0.2 + 3y0 
0.00/0.18	  filter = \G0y1.3 + 3y1 + G0(0) 
0.00/0.18	  g = \y0.2 + 2y0 
0.00/0.18	  nil = 0 
0.00/0.18	
0.00/0.18	Using this interpretation, the requirements translate to:
0.00/0.18	
0.00/0.18	  [[f(f(_x0))]] = 8 + 9x0 > 6 + 6x0 = [[g(f(_x0))]] 
0.00/0.18	  [[g(g(_x0))]] = 6 + 4x0 > 2 + 3x0 = [[f(_x0)]] 
0.00/0.18	  [[filter(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.18	
0.00/0.18	We can thus remove the following rules:
0.00/0.18	
0.00/0.18	  f(f(X)) => g(f(X))

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