Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270215

details

property	value
status	complete
benchmark	kop12lmcs2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n012.star.cs.uiowa.edu
space	Kop_13

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.14793 seconds
cpu usage	0.148107
user time	0.130677
system time	0.01743
max virtual memory	113176.0
max residence set size	6176.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.14	YES
0.00/0.14	We consider the system theBenchmark.
0.00/0.14	
0.00/0.14	  Alphabet:
0.00/0.14	
0.00/0.14	    cons : [nat -> nat * funlist] --> funlist 
0.00/0.14	    false : [] --> bool 
0.00/0.14	    head : [funlist] --> nat -> nat 
0.00/0.14	    if : [bool * nat -> string * nat -> string] --> nat -> string 
0.00/0.14	    nil : [] --> funlist 
0.00/0.14	    s : [nat] --> nat 
0.00/0.14	    tail : [funlist] --> funlist 
0.00/0.14	    test : [nat -> nat] --> bool 
0.00/0.14	    true : [] --> bool 
0.00/0.14	
0.00/0.14	  Rules:
0.00/0.14	
0.00/0.14	    if(true, f, g) => f 
0.00/0.14	    if(false, f, g) => g 
0.00/0.14	    test(/\x.s(x)) => true 
0.00/0.14	    head(cons(f, x)) => f 
0.00/0.14	    tail(cons(f, x)) => x 
0.00/0.14	
0.00/0.14	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.14	
0.00/0.14	We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).
0.00/0.14	
0.00/0.14	In order to do so, we start by eta-expanding the system, which gives:
0.00/0.14	
0.00/0.14	  if(true, F, G, X) => F X 
0.00/0.14	  if(false, F, G, X) => G X 
0.00/0.14	  test(/\x.s(x)) => true 
0.00/0.14	  head(cons(F, X), Y) => F Y 
0.00/0.14	  tail(cons(F, X)) => X 
0.00/0.14	
0.00/0.14	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
0.00/0.14	
0.00/0.14	  Dependency Pairs P_0:
0.00/0.14	
0.00/0.14	
0.00/0.14	  Rules R_0:
0.00/0.14	
0.00/0.14	    if(true, F, G, X) => F X 
0.00/0.14	    if(false, F, G, X) => G X 
0.00/0.14	    test(/\x.s(x)) => true 
0.00/0.14	    head(cons(F, X), Y) => F Y 
0.00/0.14	    tail(cons(F, X)) => X 
0.00/0.14	
0.00/0.14	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
0.00/0.14	
0.00/0.14	We consider the dependency pair problem (P_0, R_0, static, formative).
0.00/0.14	
0.00/0.14	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.14	
0.00/0.14	
0.00/0.14	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
0.00/0.14	
0.00/0.14	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.14	
0.00/0.14	
0.00/0.14	+++ Citations +++
0.00/0.14	
0.00/0.14	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.14	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.14	[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.
0.00/0.14	EOF

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