1.80/1.37	YES
1.80/1.38	We consider the system theBenchmark.
1.80/1.38	
1.80/1.38	  Alphabet:
1.80/1.38	
1.80/1.38	    0 : [] --> a 
1.80/1.38	    1 : [] --> a 
1.80/1.38	    cons : [c * d] --> d 
1.80/1.38	    f : [a * a * a] --> a 
1.80/1.38	    false : [] --> b 
1.80/1.38	    filter : [c -> b * d] --> d 
1.80/1.38	    filter2 : [b * c -> b * c * d] --> d 
1.80/1.38	    map : [c -> c * d] --> d 
1.80/1.38	    nil : [] --> d 
1.80/1.38	    s : [a] --> a 
1.80/1.38	    true : [] --> b 
1.80/1.38	
1.80/1.38	  Rules:
1.80/1.38	
1.80/1.38	    f(0, 1, x) => f(s(x), x, x) 
1.80/1.38	    f(x, y, s(z)) => s(f(0, 1, z)) 
1.80/1.38	    map(g, nil) => nil 
1.80/1.38	    map(g, cons(x, y)) => cons(g x, map(g, y)) 
1.80/1.38	    filter(g, nil) => nil 
1.80/1.38	    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
1.80/1.38	    filter2(true, g, x, y) => cons(x, filter(g, y)) 
1.80/1.38	    filter2(false, g, x, y) => filter(g, y) 
1.80/1.38	
1.80/1.38	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
1.80/1.38	
1.80/1.38	We observe that the rules contain a first-order subset:
1.80/1.38	
1.80/1.38	  f(0, 1, X) => f(s(X), X, X) 
1.80/1.38	  f(X, Y, s(Z)) => s(f(0, 1, Z)) 
1.80/1.38	
1.80/1.38	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.
1.80/1.38	
1.80/1.38	According to the external first-order termination prover, this system is indeed Ce-terminating:
1.80/1.38	
1.80/1.38	 || proof of resources/system.trs
1.80/1.38	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
1.80/1.38	 || 
1.80/1.38	 || 
1.80/1.38	 || Termination w.r.t. Q of the given QTRS could be proven:
1.80/1.38	 || 
1.80/1.38	 || (0) QTRS
1.80/1.38	 || (1) DependencyPairsProof [EQUIVALENT]
1.80/1.38	 || (2) QDP
1.80/1.38	 || (3) QDPSizeChangeProof [EQUIVALENT]
1.80/1.38	 || (4) YES
1.80/1.38	 || 
1.80/1.38	 || 
1.80/1.38	 || ----------------------------------------
1.80/1.38	 || 
1.80/1.38	 || (0)
1.80/1.38	 || Obligation:
1.80/1.38	 || Q restricted rewrite system:
1.80/1.38	 || The TRS R consists of the following rules:
1.80/1.38	 || 
1.80/1.38	 ||    f(0, 1, %X) -> f(s(%X), %X, %X)
1.80/1.38	 ||    f(%X, %Y, s(%Z)) -> s(f(0, 1, %Z))
1.80/1.38	 ||    ~PAIR(%X, %Y) -> %X
1.80/1.38	 ||    ~PAIR(%X, %Y) -> %Y
1.80/1.38	 || 
1.80/1.38	 || Q is empty.
1.80/1.38	 || 
1.80/1.38	 || ----------------------------------------
1.80/1.38	 || 
1.80/1.38	 || (1) DependencyPairsProof (EQUIVALENT)
1.80/1.38	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
1.80/1.38	 || ----------------------------------------
1.80/1.38	 || 
1.80/1.38	 || (2)
1.80/1.38	 || Obligation:
1.80/1.38	 || Q DP problem:
1.80/1.38	 || The TRS P consists of the following rules:
1.80/1.38	 || 
1.80/1.38	 ||    F(0, 1, %X) -> F(s(%X), %X, %X)
1.80/1.38	 ||    F(%X, %Y, s(%Z)) -> F(0, 1, %Z)
1.80/1.38	 || 
1.80/1.38	 || The TRS R consists of the following rules:
1.80/1.38	 || 
1.80/1.38	 ||    f(0, 1, %X) -> f(s(%X), %X, %X)
1.80/1.38	 ||    f(%X, %Y, s(%Z)) -> s(f(0, 1, %Z))
1.80/1.38	 ||    ~PAIR(%X, %Y) -> %X
1.80/1.38	 ||    ~PAIR(%X, %Y) -> %Y
1.80/1.38	 || 
1.80/1.38	 || Q is empty.
1.80/1.38	 || We have to consider all minimal (P,Q,R)-chains.
1.80/1.38	 || ----------------------------------------
1.80/1.38	 || 
1.80/1.38	 || (3) QDPSizeChangeProof (EQUIVALENT)
1.80/1.38	 || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
1.80/1.38	 || 
1.80/1.38	 || From the DPs we obtained the following set of size-change graphs:
1.80/1.38	 || *F(%X, %Y, s(%Z)) -> F(0, 1, %Z)
1.80/1.38	 || The graph contains the following edges 3 > 3
1.80/1.38	 || 
1.80/1.38	 || 
1.80/1.38	 || *F(0, 1, %X) -> F(s(%X), %X, %X)
1.80/1.38	 || The graph contains the following edges 3 >= 2, 3 >= 3
1.80/1.38	 || 
1.80/1.38	 || 
1.80/1.38	 || ----------------------------------------
1.80/1.38	 || 
1.80/1.38	 || (4)
1.80/1.38	 || YES
1.80/1.38	 || 
1.80/1.38	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]).
1.80/1.38	
1.80/1.38	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
1.80/1.38	
1.80/1.38	  Dependency Pairs P_0:
1.80/1.38	
1.80/1.38	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
1.80/1.38	    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
1.80/1.38	    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	
1.80/1.38	  Rules R_0:
1.80/1.38	
1.80/1.38	    f(0, 1, X) => f(s(X), X, X) 
1.80/1.38	    f(X, Y, s(Z)) => s(f(0, 1, Z)) 
1.80/1.38	    map(F, nil) => nil 
1.80/1.38	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
1.80/1.38	    filter(F, nil) => nil 
1.80/1.38	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
1.80/1.38	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
1.80/1.38	    filter2(false, F, X, Y) => filter(F, Y) 
1.80/1.38	
1.80/1.38	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
1.80/1.38	
1.80/1.38	We consider the dependency pair problem (P_0, R_0, static, formative).
1.80/1.38	
1.80/1.38	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
1.80/1.38	
1.80/1.38	    * 0 : 0 
1.80/1.38	    * 1 : 2, 3 
1.80/1.38	    * 2 : 1 
1.80/1.38	    * 3 : 1 
1.80/1.38	
1.80/1.38	This graph has the following strongly connected components:
1.80/1.38	
1.80/1.38	  P_1:
1.80/1.38	
1.80/1.38	    map#(F, cons(X, Y)) =#> map#(F, Y)   
1.80/1.38	
1.80/1.38	  P_2:
1.80/1.38	
1.80/1.38	    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
1.80/1.38	    filter2#(true, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	    filter2#(false, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	
1.80/1.38	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) and (P_2, R_0, m, f).
1.80/1.38	
1.80/1.38	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
1.80/1.38	
1.80/1.38	We consider the dependency pair problem (P_2, R_0, static, formative).
1.80/1.38	
1.80/1.38	We apply the subterm criterion with the following projection function:
1.80/1.38	
1.80/1.38	  nu(filter2#) = 4 
1.80/1.38	  nu(filter#) = 2 
1.80/1.38	
1.80/1.38	Thus, we can orient the dependency pairs as follows:
1.80/1.38	
1.80/1.38	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
1.80/1.38	  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
1.80/1.38	  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
1.80/1.38	
1.80/1.38	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by (P_3, R_0, static, f), where P_3 contains:
1.80/1.38	
1.80/1.38	  filter2#(true, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	  filter2#(false, F, X, Y) =#> filter#(F, Y)   
1.80/1.38	
1.80/1.38	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
1.80/1.38	
1.80/1.38	We consider the dependency pair problem (P_3, R_0, static, formative).
1.80/1.38	
1.80/1.38	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
1.80/1.38	
1.80/1.38	    * 0 :  
1.80/1.38	    * 1 :  
1.80/1.38	
1.80/1.38	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
1.80/1.38	
1.80/1.38	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
1.80/1.38	
1.80/1.38	We consider the dependency pair problem (P_1, R_0, static, formative).
1.80/1.38	
1.80/1.38	We apply the subterm criterion with the following projection function:
1.80/1.38	
1.80/1.38	  nu(map#) = 2 
1.80/1.38	
1.80/1.38	Thus, we can orient the dependency pairs as follows:
1.80/1.38	
1.80/1.38	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
1.80/1.38	
1.80/1.38	By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.
1.80/1.38	
1.80/1.38	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
1.80/1.38	
1.80/1.38	
1.80/1.38	+++ Citations +++
1.80/1.38	
1.80/1.38	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
1.80/1.38	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
1.80/1.38	[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.
1.80/1.38	EOF