Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270251

details

property	value
status	complete
benchmark	twice_modif4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n052.star.cs.uiowa.edu
space	Mixed_HO_12

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.096126 seconds
cpu usage	0.092214
user time	0.067078
system time	0.025136
max virtual memory	113176.0
max residence set size	4020.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.09	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 : [] --> nat 
0.00/0.09	    I : [nat] --> nat 
0.00/0.09	    s : [nat] --> nat 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    I(s(x)) => s((/\f./\y.f (f y)) (/\z.I(z)) x) 
0.00/0.09	    I(0) => 0 
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 the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.
0.00/0.09	
0.00/0.09	We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):
0.00/0.09	
0.00/0.09	  Dependency Pairs P_0:
0.00/0.09	
0.00/0.09	    0] I#(s(X)) =#> I#(x)   
0.00/0.09	    1] I#(s(X)) =#> I#(X)   
0.00/0.09	    2] I#(s(X)) =#> I#((/\x.I(x)) X)   
0.00/0.09	    3] I#(s(X)) =#> I#(X)   
0.00/0.09	
0.00/0.09	  Rules R_0:
0.00/0.09	
0.00/0.09	    I(s(X)) => s((/\f./\x.f (f x)) (/\y.I(y)) X) 
0.00/0.09	    I(0) => 0 
0.00/0.09	
0.00/0.09	Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.
0.00/0.09	
0.00/0.09	We consider the dependency pair problem (P_0, R_0, minimal, formative).
0.00/0.09	
0.00/0.09	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.09	
0.00/0.09	    * 0 :  
0.00/0.09	    * 1 : 0, 1, 2, 3 
0.00/0.09	    * 2 : 0, 1, 2, 3 
0.00/0.09	    * 3 : 0, 1, 2, 3 
0.00/0.09	
0.00/0.09	This graph has the following strongly connected components:
0.00/0.09	
0.00/0.09	  P_1:
0.00/0.09	
0.00/0.09	    I#(s(X)) =#> I#(X)   
0.00/0.09	    I#(s(X)) =#> I#((/\x.I(x)) X)   
0.00/0.09	    I#(s(X)) =#> I#(X)   
0.00/0.09	
0.00/0.09	By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).
0.00/0.09	
0.00/0.09	Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.
0.00/0.09	
0.00/0.09	We consider the dependency pair problem (P_1, R_0, minimal, formative).
0.00/0.09	
0.00/0.09	The formative rules of (P_1, R_0) are R_1 ::=
0.00/0.09	
0.00/0.09	  I(s(X)) => s((/\f./\x.f (f x)) (/\y.I(y)) X) 
0.00/0.09	
0.00/0.09	By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).
0.00/0.09	
0.00/0.09	Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.
0.00/0.09	
0.00/0.09	We consider the dependency pair problem (P_1, R_1, minimal, formative).
0.00/0.09	
0.00/0.09	We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:
0.00/0.09	
0.00/0.09	  I#(s(X)) >? I#(X) 
0.00/0.09	  I#(s(X)) >? I#((/\x.I(x)) X) 
0.00/0.09	  I#(s(X)) >? I#(X) 
0.00/0.09	  I(s(X)) >= s((/\f./\x.f (f x)) (/\y.I(y)) X) 
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	  I = \y0.y0 
0.00/0.09	  I# = \y0.3y0 
0.00/0.09	  s = \y0.3 + 3y0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#(_x0)]] 
0.00/0.09	  [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#((/\x.I(x)) _x0)]] 
0.00/0.09	  [[I#(s(_x0))]] = 9 + 9x0 > 3x0 = [[I#(_x0)]] 
0.00/0.09	  [[I(s(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[s((/\f./\x.f (f x)) (/\y.I(y)) _x0)]] 
0.00/0.09	
0.00/0.09	By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, R_1) by ({}, R_1).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
0.00/0.09	
0.00/0.09	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
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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