1.83/1.10	YES
1.83/1.11	We consider the system theBenchmark.
1.83/1.11	
1.83/1.11	  Alphabet:
1.83/1.11	
1.83/1.11	    0 : [] --> N 
1.83/1.11	    false : [] --> B 
1.83/1.11	    g : [] --> N -> B 
1.83/1.11	    h : [] --> N -> (N -> B) -> N -> B 
1.83/1.11	    iszero : [] --> N -> N -> B 
1.83/1.11	    rec : [] --> (N -> (N -> B) -> N -> B) -> B -> N -> B 
1.83/1.11	    true : [] --> B 
1.83/1.11	
1.83/1.11	  Rules:
1.83/1.11	
1.83/1.11	    rec f (i 0) => i 
1.83/1.11	    g x => true 
1.83/1.11	    h x f y => false 
1.83/1.11	    iszero x y => rec h (g x) y 
1.83/1.11	
1.83/1.11	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.
1.83/1.11	
1.83/1.11	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:
1.83/1.11	
1.83/1.11	  Alphabet:
1.83/1.11	
1.83/1.11	    0 : [] --> N 
1.83/1.11	    false : [] --> B 
1.83/1.11	    g : [] --> N -> B 
1.83/1.11	    h : [] --> N -> (N -> B) -> N -> B 
1.83/1.11	    iszero : [N] --> N -> B 
1.83/1.11	    rec : [N -> (N -> B) -> N -> B * B] --> N -> B 
1.83/1.11	    true : [] --> B 
1.83/1.11	    ~AP1 : [N -> B * N] --> B 
1.83/1.11	
1.83/1.11	  Rules:
1.83/1.11	
1.83/1.11	    rec(F, ~AP1(G, 0)) => G 
1.83/1.11	    g X => true 
1.83/1.11	    h X F Y => false 
1.83/1.11	    iszero(X) Y => ~AP1(rec(h, g X), Y) 
1.83/1.11	    rec(F, g 0) => g 
1.83/1.11	    rec(F, h X G 0) => h X G 
1.83/1.11	    rec(F, iszero(X) 0) => iszero(X) 
1.83/1.11	    ~AP1(F, X) => F X 
1.83/1.11	
1.83/1.11	We observe that the rules contain a first-order subset:
1.83/1.11	
1.83/1.11	  g X => true 
1.83/1.11	
1.83/1.11	Moreover, the system is finitely branching.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS.
1.83/1.11	
1.83/1.11	According to the external first-order termination prover, this system is indeed Ce-terminating:
1.83/1.11	
1.83/1.11	 || proof of resources/system.trs
1.83/1.11	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
1.83/1.11	 || 
1.83/1.11	 || 
1.83/1.11	 || Termination w.r.t. Q of the given QTRS could be proven:
1.83/1.11	 || 
1.83/1.11	 || (0) QTRS
1.83/1.11	 || (1) QTRSRRRProof [EQUIVALENT]
1.83/1.11	 || (2) QTRS
1.83/1.11	 || (3) RisEmptyProof [EQUIVALENT]
1.83/1.11	 || (4) YES
1.83/1.11	 || 
1.83/1.11	 || 
1.83/1.11	 || ----------------------------------------
1.83/1.11	 || 
1.83/1.11	 || (0)
1.83/1.11	 || Obligation:
1.83/1.11	 || Q restricted rewrite system:
1.83/1.11	 || The TRS R consists of the following rules:
1.83/1.11	 || 
1.83/1.11	 ||    g(%X) -> true
1.83/1.11	 ||    ~PAIR(%X, %Y) -> %X
1.83/1.11	 ||    ~PAIR(%X, %Y) -> %Y
1.83/1.11	 || 
1.83/1.11	 || Q is empty.
1.83/1.11	 || 
1.83/1.11	 || ----------------------------------------
1.83/1.11	 || 
1.83/1.11	 || (1) QTRSRRRProof (EQUIVALENT)
1.83/1.11	 || Used ordering:
1.83/1.11	 || Polynomial interpretation [POLO]:
1.83/1.11	 || 
1.83/1.11	 ||    POL(g(x_1)) = 2 + x_1
1.83/1.11	 ||    POL(true) = 1
1.83/1.11	 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
1.83/1.11	 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
1.83/1.11	 || 
1.83/1.11	 ||    g(%X) -> true
1.83/1.11	 ||    ~PAIR(%X, %Y) -> %X
1.83/1.11	 ||    ~PAIR(%X, %Y) -> %Y
1.83/1.11	 || 
1.83/1.11	 || 
1.83/1.11	 || 
1.83/1.11	 || 
1.83/1.11	 || ----------------------------------------
1.83/1.11	 || 
1.83/1.11	 || (2)
1.83/1.11	 || Obligation:
1.83/1.11	 || Q restricted rewrite system:
1.83/1.11	 || R is empty.
1.83/1.11	 || Q is empty.
1.83/1.11	 || 
1.83/1.11	 || ----------------------------------------
1.83/1.11	 || 
1.83/1.11	 || (3) RisEmptyProof (EQUIVALENT)
1.83/1.11	 || The TRS R is empty. Hence, termination is trivially proven.
1.83/1.11	 || ----------------------------------------
1.83/1.11	 || 
1.83/1.11	 || (4)
1.83/1.11	 || YES
1.83/1.11	 || 
1.83/1.11	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]).
1.83/1.11	
1.83/1.11	In order to do so, we start by eta-expanding the system, which gives:
1.83/1.11	
1.83/1.11	  rec(F, ~AP1(G, 0), X) => G X 
1.83/1.11	  g(X) => true 
1.83/1.11	  h(X, F, Y) => false 
1.83/1.11	  iszero(X, Y) => ~AP1(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y) 
1.83/1.11	  rec(F, g(0), X) => g(X) 
1.83/1.11	  rec(F, h(X, G, 0), Y) => h(X, G, Y) 
1.83/1.11	  rec(F, iszero(X, 0), Y) => iszero(X, Y) 
1.83/1.11	  ~AP1(F, X) => F X 
1.83/1.11	
1.83/1.11	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
1.83/1.11	
1.83/1.11	  Dependency Pairs P_0:
1.83/1.11	
1.83/1.11	    0] iszero#(X, Y) =#> ~AP1#(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y)   
1.83/1.11	    1] iszero#(X, Y) =#> rec#(/\x./\f./\y.h(x, /\z.f z, y), g(X), Z)   
1.83/1.11	    2] iszero#(X, Y) =#> h#(Z, /\x.F x, U)   
1.83/1.11	    3] iszero#(X, Y) =#> g#(X)   
1.83/1.11	    4] rec#(F, g(0), X) =#> g#(X)   
1.83/1.11	    5] rec#(F, h(X, G, 0), Y) =#> h#(X, G, Y)   
1.83/1.11	    6] rec#(F, iszero(X, 0), Y) =#> iszero#(X, Y)   
1.83/1.11	
1.83/1.11	  Rules R_0:
1.83/1.11	
1.83/1.11	    rec(F, ~AP1(G, 0), X) => G X 
1.83/1.11	    g(X) => true 
1.83/1.11	    h(X, F, Y) => false 
1.83/1.11	    iszero(X, Y) => ~AP1(/\x.rec(/\y./\f./\z.h(y, /\u.f u, z), g(X), x), Y) 
1.83/1.11	    rec(F, g(0), X) => g(X) 
1.83/1.11	    rec(F, h(X, G, 0), Y) => h(X, G, Y) 
1.83/1.11	    rec(F, iszero(X, 0), Y) => iszero(X, Y) 
1.83/1.11	    ~AP1(F, X) => F X 
1.83/1.11	
1.83/1.11	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
1.83/1.11	
1.83/1.11	We consider the dependency pair problem (P_0, R_0, static, formative).
1.83/1.11	
1.83/1.11	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
1.83/1.11	
1.83/1.11	    * 0 :  
1.83/1.11	    * 1 : 4 
1.83/1.11	    * 2 :  
1.83/1.11	    * 3 :  
1.83/1.11	    * 4 :  
1.83/1.11	    * 5 :  
1.83/1.11	    * 6 : 0, 1, 2, 3 
1.83/1.11	
1.83/1.11	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
1.83/1.11	
1.83/1.11	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
1.83/1.11	
1.83/1.11	
1.83/1.11	+++ Citations +++
1.83/1.11	
1.83/1.11	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
1.83/1.11	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
1.83/1.11	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
1.83/1.11	[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.
1.83/1.11	EOF