Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270355

details

property	value
status	complete
benchmark	SystemT.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n109.star.cs.uiowa.edu
space	Hamana_17

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.106117 seconds
cpu usage	0.106223
user time	0.090781
system time	0.015442
max virtual memory	113176.0
max residence set size	4840.0

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	    app : [] --> arrab -> a -> b 
0.00/0.10	    lam : [] --> (a -> b) -> arrab 
0.00/0.10	    rec : [] --> Nat -> a -> (Nat -> a -> a) -> a 
0.00/0.10	    succ : [] --> Nat -> Nat 
0.00/0.10	    zero : [] --> Nat 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app (lam (/\x.f x)) y => f y 
0.00/0.10	    lam (/\x.app y x) => y 
0.00/0.10	    rec zero x (/\y.f y) => x 
0.00/0.10	    rec (succ x) y (/\z.f z) => f x (rec x y (/\u.f u)) 
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	    app : [arrab * a] --> b 
0.00/0.10	    lam : [a -> b] --> arrab 
0.00/0.10	    rec : [Nat * a * Nat -> a -> a] --> a 
0.00/0.10	    succ : [Nat] --> Nat 
0.00/0.10	    zero : [] --> Nat 
0.00/0.10	    ~AP1 : [a -> b * a] --> b 
0.00/0.10	    ~AP2 : [Nat -> a -> a * Nat] --> a -> a 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app(lam(/\x.~AP1(F, x)), X) => ~AP1(F, X) 
0.00/0.10	    lam(/\x.app(X, x)) => X 
0.00/0.10	    rec(zero, X, /\x.~AP2(F, x)) => X 
0.00/0.10	    rec(succ(X), Y, /\x.~AP2(F, x)) => ~AP2(F, X) rec(X, Y, /\y.~AP2(F, y)) 
0.00/0.10	    app(lam(/\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	    app : [arrab * a] --> b 
0.00/0.10	    lam : [a -> b] --> arrab 
0.00/0.10	    rec : [Nat * a * Nat -> a -> a] --> a 
0.00/0.10	    succ : [Nat] --> Nat 
0.00/0.10	    zero : [] --> Nat 
0.00/0.10	
0.00/0.10	  Rules:
0.00/0.10	
0.00/0.10	    app(lam(/\x.X(x)), Y) => X(Y) 
0.00/0.10	    lam(/\x.app(X, x)) => X 
0.00/0.10	    rec(zero, X, /\x.F(x)) => X 
0.00/0.10	    rec(succ(X), Y, /\x.F(x)) => F(X) rec(X, Y, /\y.F(y)) 
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] rec#(succ(X), Y, /\x.F(x)) =#> rec#(X, Y, /\y.F(y))   
0.00/0.10	
0.00/0.10	  Rules R_0:
0.00/0.10	
0.00/0.10	    app(lam(/\x.X(x)), Y) => X(Y) 
0.00/0.10	    lam(/\x.app(X, x)) => X 
0.00/0.10	    rec(zero, X, /\x.F(x)) => X 
0.00/0.10	    rec(succ(X), Y, /\x.F(x)) => F(X) rec(X, Y, /\y.F(y)) 
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 apply the subterm criterion with the following projection function:
0.00/0.10	
0.00/0.10	  nu(rec#) = 1 
0.00/0.10	
0.00/0.10	Thus, we can orient the dependency pairs as follows:
0.00/0.10	
0.00/0.10	  nu(rec#(succ(X), Y, /\x.F(x))) = succ(X) |> X = nu(rec#(X, Y, /\y.F(y))) 
0.00/0.10	
0.00/0.10	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_0, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
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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