Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270553

details

property	value
status	complete
benchmark	AotoYamada_05__027.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n150.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.0907291 seconds
cpu usage	0.09086
user time	0.07725
system time	0.01361
max virtual memory	113176.0
max residence set size	2860.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.08	YES
0.00/0.09	We consider the system theBenchmark.
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    0 : [] --> a 
0.00/0.09	    cons : [a * b] --> b 
0.00/0.09	    inc : [b] --> b 
0.00/0.09	    map : [a -> a * b] --> b 
0.00/0.09	    nil : [] --> b 
0.00/0.09	    plus : [a] --> a -> a 
0.00/0.09	    s : [a] --> a 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    plus(0) x => x 
0.00/0.09	    plus(s(x)) y => s(plus(x) y) 
0.00/0.09	    inc(x) => map(plus(s(0)), x) 
0.00/0.09	    map(f, nil) => nil 
0.00/0.09	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
0.00/0.09	
0.00/0.09	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.09	
0.00/0.09	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.09	
0.00/0.09	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.09	
0.00/0.09	  plus(0) X >? X 
0.00/0.09	  plus(s(X)) Y >? s(plus(X) Y) 
0.00/0.09	  inc(X) >? map(plus(s(0)), X) 
0.00/0.09	  map(F, nil) >? nil 
0.00/0.09	  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
0.00/0.09	
0.00/0.09	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.09	
0.00/0.09	The following interpretation satisfies the requirements:
0.00/0.09	
0.00/0.09	  0 = 0 
0.00/0.09	  cons = \y0y1.3 + y1 + 2y0 
0.00/0.09	  inc = \y0.3 + 3y0 
0.00/0.09	  map = \G0y1.2 + y1 + G0(0) + 3y1G0(y1) 
0.00/0.09	  nil = 0 
0.00/0.09	  plus = \y0y1.2y0 
0.00/0.09	  s = \y0.y0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[plus(0) _x0]] = x0 >= x0 = [[_x0]] 
0.00/0.09	  [[plus(s(_x0)) _x1]] = x1 + 2x0 >= x1 + 2x0 = [[s(plus(_x0) _x1)]] 
0.00/0.09	  [[inc(_x0)]] = 3 + 3x0 > 2 + x0 = [[map(plus(s(0)), _x0)]] 
0.00/0.09	  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
0.00/0.09	  [[map(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 5 + x2 + 2x1 + F0(0) + 2F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  inc(X) => map(plus(s(0)), X) 
0.00/0.09	  map(F, nil) => nil 
0.00/0.09	
0.00/0.09	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.09	
0.00/0.09	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.09	
0.00/0.09	  plus(0, X) >? X 
0.00/0.09	  plus(s(X), Y) >? s(plus(X, Y)) 
0.00/0.09	  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
0.00/0.09	
0.00/0.09	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.09	
0.00/0.09	The following interpretation satisfies the requirements:
0.00/0.09	
0.00/0.09	  0 = 3 
0.00/0.09	  cons = \y0y1.3 + y0 + y1 
0.00/0.09	  map = \G0y1.3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) 
0.00/0.09	  plus = \y0y1.3 + y1 + 3y0 
0.00/0.09	  s = \y0.3 + y0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[plus(0, _x0)]] = 12 + x0 > x0 = [[_x0]] 
0.00/0.09	  [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
0.00/0.09	  [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2F0(0) + 2F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  plus(0, X) => X 
0.00/0.09	  plus(s(X), Y) => s(plus(X, Y)) 
0.00/0.09	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.09	
0.00/0.09	All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.
0.00/0.09	
0.00/0.09	
0.00/0.09	+++ Citations +++
0.00/0.09	
0.00/0.09	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.09	EOF

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