Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270205

details

property	value
status	complete
benchmark	fuhkop12rta1.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n008.star.cs.uiowa.edu
space	Kop_13

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.178224 seconds
cpu usage	0.178608
user time	0.146756
system time	0.031852
max virtual memory	113176.0
max residence set size	3712.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.16	YES
0.00/0.17	We consider the system theBenchmark.
0.00/0.17	
0.00/0.17	  Alphabet:
0.00/0.17	
0.00/0.17	    app : [list * list] --> list 
0.00/0.17	    cons : [nat * list] --> list 
0.00/0.17	    hshuffle : [nat -> nat * list] --> list 
0.00/0.17	    nil : [] --> list 
0.00/0.17	    reverse : [list] --> list 
0.00/0.17	
0.00/0.17	  Rules:
0.00/0.17	
0.00/0.17	    app(nil, x) => x 
0.00/0.17	    app(cons(x, y), z) => cons(x, app(y, z)) 
0.00/0.17	    reverse(nil) => nil 
0.00/0.17	    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
0.00/0.17	    hshuffle(f, nil) => nil 
0.00/0.17	    hshuffle(f, cons(x, y)) => cons(f x, hshuffle(f, reverse(y))) 
0.00/0.17	
0.00/0.17	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.17	
0.00/0.17	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.17	
0.00/0.17	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.17	
0.00/0.17	  app(nil, X) >? X 
0.00/0.17	  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
0.00/0.17	  reverse(nil) >? nil 
0.00/0.17	  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
0.00/0.17	  hshuffle(F, nil) >? nil 
0.00/0.17	  hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 
0.00/0.17	
0.00/0.17	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.17	
0.00/0.17	The following interpretation satisfies the requirements:
0.00/0.17	
0.00/0.17	  app = \y0y1.y0 + y1 
0.00/0.17	  cons = \y0y1.2 + y0 + y1 
0.00/0.17	  hshuffle = \G0y1.2y1 + G0(0) + y1G0(y1) 
0.00/0.17	  nil = 0 
0.00/0.17	  reverse = \y0.1 + y0 
0.00/0.17	
0.00/0.17	Using this interpretation, the requirements translate to:
0.00/0.17	
0.00/0.17	  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
0.00/0.17	  [[app(cons(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
0.00/0.17	  [[reverse(nil)]] = 1 > 0 = [[nil]] 
0.00/0.17	  [[reverse(cons(_x0, _x1))]] = 3 + x0 + x1 >= 3 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
0.00/0.17	  [[hshuffle(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
0.00/0.17	  [[hshuffle(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 4 + x1 + 2x2 + F0(0) + F0(x1) + F0(1 + x2) + x2F0(1 + x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 
0.00/0.17	
0.00/0.17	We can thus remove the following rules:
0.00/0.17	
0.00/0.17	  reverse(nil) => nil 
0.00/0.17	
0.00/0.17	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.17	
0.00/0.17	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.17	
0.00/0.17	  app(nil, X) >? X 
0.00/0.17	  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
0.00/0.17	  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
0.00/0.17	  hshuffle(F, nil) >? nil 
0.00/0.17	  hshuffle(F, cons(X, Y)) >? cons(F X, hshuffle(F, reverse(Y))) 
0.00/0.17	
0.00/0.17	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.17	
0.00/0.17	The following interpretation satisfies the requirements:
0.00/0.17	
0.00/0.17	  app = \y0y1.y0 + y1 
0.00/0.17	  cons = \y0y1.1 + y0 + y1 
0.00/0.17	  hshuffle = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 
0.00/0.17	  nil = 0 
0.00/0.17	  reverse = \y0.y0 
0.00/0.17	
0.00/0.17	Using this interpretation, the requirements translate to:
0.00/0.17	
0.00/0.17	  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
0.00/0.17	  [[app(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
0.00/0.17	  [[reverse(cons(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
0.00/0.17	  [[hshuffle(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
0.00/0.17	  [[hshuffle(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > 1 + x1 + 2x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 
0.00/0.17	
0.00/0.17	We can thus remove the following rules:
0.00/0.17	
0.00/0.17	  hshuffle(F, cons(X, Y)) => cons(F X, hshuffle(F, reverse(Y))) 
0.00/0.17	
0.00/0.17	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.17	
0.00/0.17	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.17	
0.00/0.17	  app(nil, X) >? X 
0.00/0.17	  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
0.00/0.17	  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
0.00/0.17	  hshuffle(F, nil) >? nil 
0.00/0.17	
0.00/0.17	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.17	
0.00/0.17	The following interpretation satisfies the requirements:

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