Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270361

details

property	value
status	complete
benchmark	ysllc.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n006.star.cs.uiowa.edu
space	Hamana_17

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.108532 seconds
cpu usage	0.10856
user time	0.095183
system time	0.013377
max virtual memory	113176.0
max residence set size	5900.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.10	YES
0.00/0.10	We consider the system theBenchmark.
0.00/0.10	
0.00/0.10	  Alphabet:
0.00/0.10	
0.00/0.10	    abs : [] --> (a -> b) -> ab 
0.00/0.10	    app : [] --> ab -> a -> b 
0.00/0.10	    appBG : [] --> BGaBGb -> BGa -> BGb 
0.00/0.10	    banga : [] --> a -> BGa 
0.00/0.10	    let : [] --> BGa -> (a -> BGb) -> BGb 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app (abs (/\x.f x)) y => f y 
0.00/0.10	    abs (/\x.app y x) => y 
0.00/0.10	    let (banga x) (/\y.f y) => f x 
0.00/0.10	    let x (/\y.appBG z (banga y)) => appBG z x 
0.00/0.10	
0.00/0.10	Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.
0.00/0.10	
0.00/0.10	We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:
0.00/0.10	
0.00/0.10	  Alphabet:
0.00/0.10	
0.00/0.10	    abs : [a -> b] --> ab 
0.00/0.10	    app : [ab * a] --> b 
0.00/0.10	    appBG : [BGaBGb * BGa] --> BGb 
0.00/0.10	    banga : [a] --> BGa 
0.00/0.10	    let : [BGa * a -> BGb] --> BGb 
0.00/0.10	    ~AP1 : [a -> b * a] --> b 
0.00/0.10	    ~AP2 : [a -> BGb * a] --> BGb 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app(abs(/\x.~AP1(F, x)), X) => ~AP1(F, X) 
0.00/0.10	    abs(/\x.app(X, x)) => X 
0.00/0.10	    let(banga(X), /\x.~AP2(F, x)) => ~AP2(F, X) 
0.00/0.10	    let(X, /\x.appBG(Y, banga(x))) => appBG(Y, X) 
0.00/0.10	    app(abs(/\x.app(X, x)), Y) => app(X, Y) 
0.00/0.10	    ~AP1(F, X) => F X 
0.00/0.10	    ~AP2(F, X) => F X 
0.00/0.10	
0.00/0.10	Symbols ~AP1, and ~AP2 are encodings for application that are only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing them.
0.00/0.10	
0.00/0.10	  Additionally, we can remove some (now-)redundant rules.  This gives:
0.00/0.10	
0.00/0.10	  Alphabet:
0.00/0.10	
0.00/0.10	    abs : [a -> b] --> ab 
0.00/0.10	    app : [ab * a] --> b 
0.00/0.10	    appBG : [BGaBGb * BGa] --> BGb 
0.00/0.10	    banga : [a] --> BGa 
0.00/0.10	    let : [BGa * a -> BGb] --> BGb 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app(abs(/\x.X(x)), Y) => X(Y) 
0.00/0.10	    abs(/\x.app(X, x)) => X 
0.00/0.10	    let(banga(X), /\x.Y(x)) => Y(X) 
0.00/0.10	    let(X, /\x.appBG(Y, banga(x))) => appBG(Y, X) 
0.00/0.10	
0.00/0.10	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]).
0.00/0.10	
0.00/0.10	We thus obtain the following dependency pair problem (P_0, R_0, static, all):
0.00/0.10	
0.00/0.10	  Dependency Pairs P_0:
0.00/0.10	
0.00/0.10	
0.00/0.10	  Rules R_0:
0.00/0.10	
0.00/0.10	    app(abs(/\x.X(x)), Y) => X(Y) 
0.00/0.10	    abs(/\x.app(X, x)) => X 
0.00/0.10	    let(banga(X), /\x.Y(x)) => Y(X) 
0.00/0.10	    let(X, /\x.appBG(Y, banga(x))) => appBG(Y, X) 
0.00/0.10	
0.00/0.10	Thus, the original system is terminating if (P_0, R_0, static, all) is finite.
0.00/0.10	
0.00/0.10	We consider the dependency pair problem (P_0, R_0, static, all).
0.00/0.10	
0.00/0.10	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.10	
0.00/0.10	
0.00/0.10	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
0.00/0.10	
0.00/0.10	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
0.00/0.10	
0.00/0.10	
0.00/0.10	+++ Citations +++
0.00/0.10	
0.00/0.10	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.00/0.10	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.10	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
0.00/0.10	[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.
0.00/0.10	EOF

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