2.47/1.27	YES
2.47/1.28	We consider the system theBenchmark.
2.47/1.28	
2.47/1.28	  Alphabet:
2.47/1.28	
2.47/1.28	    0 : [] --> a 
2.47/1.28	    cons : [c * d] --> d 
2.47/1.28	    double : [a] --> a 
2.47/1.28	    false : [] --> b 
2.47/1.28	    filter : [c -> b * d] --> d 
2.47/1.28	    filter2 : [b * c -> b * c * d] --> d 
2.47/1.28	    map : [c -> c * d] --> d 
2.47/1.28	    minus : [a * a] --> a 
2.47/1.28	    nil : [] --> d 
2.47/1.28	    plus : [a * a] --> a 
2.47/1.28	    s : [a] --> a 
2.47/1.28	    true : [] --> b 
2.47/1.28	
2.47/1.28	  Rules:
2.47/1.28	
2.47/1.28	    minus(x, 0) => x 
2.47/1.28	    minus(s(x), s(y)) => minus(x, y) 
2.47/1.28	    double(0) => 0 
2.47/1.28	    double(s(x)) => s(s(double(x))) 
2.47/1.28	    plus(0, x) => x 
2.47/1.28	    plus(s(x), y) => s(plus(x, y)) 
2.47/1.28	    plus(s(x), y) => plus(x, s(y)) 
2.47/1.28	    plus(s(x), y) => s(plus(minus(x, y), double(y))) 
2.47/1.28	    map(f, nil) => nil 
2.47/1.28	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
2.47/1.28	    filter(f, nil) => nil 
2.47/1.28	    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
2.47/1.28	    filter2(true, f, x, y) => cons(x, filter(f, y)) 
2.47/1.28	    filter2(false, f, x, y) => filter(f, y) 
2.47/1.28	
2.47/1.28	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.47/1.28	
2.47/1.28	We observe that the rules contain a first-order subset:
2.47/1.28	
2.47/1.28	  minus(X, 0) => X 
2.47/1.28	  minus(s(X), s(Y)) => minus(X, Y) 
2.47/1.28	  double(0) => 0 
2.47/1.28	  double(s(X)) => s(s(double(X))) 
2.47/1.28	  plus(0, X) => X 
2.47/1.28	  plus(s(X), Y) => s(plus(X, Y)) 
2.47/1.28	  plus(s(X), Y) => plus(X, s(Y)) 
2.47/1.28	  plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 
2.47/1.28	
2.47/1.28	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.
2.47/1.28	
2.47/1.28	According to the external first-order termination prover, this system is indeed Ce-terminating:
2.47/1.28	
2.47/1.28	 || proof of resources/system.trs
2.47/1.28	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || Termination w.r.t. Q of the given QTRS could be proven:
2.47/1.28	 || 
2.47/1.28	 || (0) QTRS
2.47/1.28	 || (1) DependencyPairsProof [EQUIVALENT]
2.47/1.28	 || (2) QDP
2.47/1.28	 || (3) DependencyGraphProof [EQUIVALENT]
2.47/1.28	 || (4) AND
2.47/1.28	 ||     (5) QDP
2.47/1.28	 ||         (6) UsableRulesProof [EQUIVALENT]
2.47/1.28	 ||         (7) QDP
2.47/1.28	 ||         (8) QDPSizeChangeProof [EQUIVALENT]
2.47/1.28	 ||         (9) YES
2.47/1.28	 ||     (10) QDP
2.47/1.28	 ||         (11) UsableRulesProof [EQUIVALENT]
2.47/1.28	 ||         (12) QDP
2.47/1.28	 ||         (13) QDPSizeChangeProof [EQUIVALENT]
2.47/1.28	 ||         (14) YES
2.47/1.28	 ||     (15) QDP
2.47/1.28	 ||         (16) QDPOrderProof [EQUIVALENT]
2.47/1.28	 ||         (17) QDP
2.47/1.28	 ||         (18) PisEmptyProof [EQUIVALENT]
2.47/1.28	 ||         (19) YES
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (0)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q restricted rewrite system:
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (1) DependencyPairsProof (EQUIVALENT)
2.47/1.28	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (2)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.47/1.28	 ||    DOUBLE(s(%X)) -> DOUBLE(%X)
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, s(%Y))
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(minus(%X, %Y), double(%Y))
2.47/1.28	 ||    PLUS(s(%X), %Y) -> MINUS(%X, %Y)
2.47/1.28	 ||    PLUS(s(%X), %Y) -> DOUBLE(%Y)
2.47/1.28	 || 
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (3) DependencyGraphProof (EQUIVALENT)
2.47/1.28	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (4)
2.47/1.28	 || Complex Obligation (AND)
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (5)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    DOUBLE(s(%X)) -> DOUBLE(%X)
2.47/1.28	 || 
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (6) UsableRulesProof (EQUIVALENT)
2.47/1.28	 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (7)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    DOUBLE(s(%X)) -> DOUBLE(%X)
2.47/1.28	 || 
2.47/1.28	 || R is empty.
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (8) QDPSizeChangeProof (EQUIVALENT)
2.47/1.28	 || 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. 
2.47/1.28	 || 
2.47/1.28	 || From the DPs we obtained the following set of size-change graphs:
2.47/1.28	 || *DOUBLE(s(%X)) -> DOUBLE(%X)
2.47/1.28	 || The graph contains the following edges 1 > 1
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (9)
2.47/1.28	 || YES
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (10)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.47/1.28	 || 
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (11) UsableRulesProof (EQUIVALENT)
2.47/1.28	 || We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (12)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.47/1.28	 || 
2.47/1.28	 || R is empty.
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (13) QDPSizeChangeProof (EQUIVALENT)
2.47/1.28	 || 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. 
2.47/1.28	 || 
2.47/1.28	 || From the DPs we obtained the following set of size-change graphs:
2.47/1.28	 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.47/1.28	 || The graph contains the following edges 1 > 1, 2 > 2
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (14)
2.47/1.28	 || YES
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (15)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || The TRS P consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, s(%Y))
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(minus(%X, %Y), double(%Y))
2.47/1.28	 || 
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (16) QDPOrderProof (EQUIVALENT)
2.47/1.28	 || We use the reduction pair processor [LPAR04,JAR06].
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || The following pairs can be oriented strictly and are deleted.
2.47/1.28	 || 
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, s(%Y))
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
2.47/1.28	 ||    PLUS(s(%X), %Y) -> PLUS(minus(%X, %Y), double(%Y))
2.47/1.28	 || The remaining pairs can at least be oriented weakly.
2.47/1.28	 || Used ordering:  Polynomial interpretation [POLO]:
2.47/1.28	 || 
2.47/1.28	 ||    POL(0) = 0
2.47/1.28	 ||    POL(PLUS(x_1, x_2)) = x_1
2.47/1.28	 ||    POL(double(x_1)) = 0
2.47/1.28	 ||    POL(minus(x_1, x_2)) = x_1
2.47/1.28	 ||    POL(s(x_1)) = 1 + x_1
2.47/1.28	 || 
2.47/1.28	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 || 
2.47/1.28	 || 
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (17)
2.47/1.28	 || Obligation:
2.47/1.28	 || Q DP problem:
2.47/1.28	 || P is empty.
2.47/1.28	 || The TRS R consists of the following rules:
2.47/1.28	 || 
2.47/1.28	 ||    minus(%X, 0) -> %X
2.47/1.28	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.47/1.28	 ||    double(0) -> 0
2.47/1.28	 ||    double(s(%X)) -> s(s(double(%X)))
2.47/1.28	 ||    plus(0, %X) -> %X
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> plus(%X, s(%Y))
2.47/1.28	 ||    plus(s(%X), %Y) -> s(plus(minus(%X, %Y), double(%Y)))
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %X
2.47/1.28	 ||    ~PAIR(%X, %Y) -> %Y
2.47/1.28	 || 
2.47/1.28	 || Q is empty.
2.47/1.28	 || We have to consider all minimal (P,Q,R)-chains.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (18) PisEmptyProof (EQUIVALENT)
2.47/1.28	 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
2.47/1.28	 || ----------------------------------------
2.47/1.28	 || 
2.47/1.28	 || (19)
2.47/1.28	 || YES
2.47/1.28	 || 
2.47/1.28	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]).
2.47/1.28	
2.47/1.28	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
2.47/1.28	
2.47/1.28	  Dependency Pairs P_0:
2.47/1.28	
2.47/1.28	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
2.47/1.28	    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
2.47/1.28	    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	
2.47/1.28	  Rules R_0:
2.47/1.28	
2.47/1.28	    minus(X, 0) => X 
2.47/1.28	    minus(s(X), s(Y)) => minus(X, Y) 
2.47/1.28	    double(0) => 0 
2.47/1.28	    double(s(X)) => s(s(double(X))) 
2.47/1.28	    plus(0, X) => X 
2.47/1.28	    plus(s(X), Y) => s(plus(X, Y)) 
2.47/1.28	    plus(s(X), Y) => plus(X, s(Y)) 
2.47/1.28	    plus(s(X), Y) => s(plus(minus(X, Y), double(Y))) 
2.47/1.28	    map(F, nil) => nil 
2.47/1.28	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
2.47/1.28	    filter(F, nil) => nil 
2.47/1.28	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
2.47/1.28	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
2.47/1.28	    filter2(false, F, X, Y) => filter(F, Y) 
2.47/1.28	
2.47/1.28	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
2.47/1.28	
2.47/1.28	We consider the dependency pair problem (P_0, R_0, static, formative).
2.47/1.28	
2.47/1.28	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.47/1.28	
2.47/1.28	    * 0 : 0 
2.47/1.28	    * 1 : 2, 3 
2.47/1.28	    * 2 : 1 
2.47/1.28	    * 3 : 1 
2.47/1.28	
2.47/1.28	This graph has the following strongly connected components:
2.47/1.28	
2.47/1.28	  P_1:
2.47/1.28	
2.47/1.28	    map#(F, cons(X, Y)) =#> map#(F, Y)   
2.47/1.28	
2.47/1.28	  P_2:
2.47/1.28	
2.47/1.28	    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
2.47/1.28	    filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	    filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	
2.47/1.28	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).
2.47/1.28	
2.47/1.28	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
2.47/1.28	
2.47/1.28	We consider the dependency pair problem (P_2, R_0, static, formative).
2.47/1.28	
2.47/1.28	We apply the subterm criterion with the following projection function:
2.47/1.28	
2.47/1.28	  nu(filter2#) = 4 
2.47/1.28	  nu(filter#) = 2 
2.47/1.28	
2.47/1.28	Thus, we can orient the dependency pairs as follows:
2.47/1.28	
2.47/1.28	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
2.47/1.28	  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
2.47/1.28	  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
2.47/1.28	
2.47/1.28	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:
2.47/1.28	
2.47/1.28	  filter2#(true, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	  filter2#(false, F, X, Y) =#> filter#(F, Y)   
2.47/1.28	
2.47/1.28	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
2.47/1.28	
2.47/1.28	We consider the dependency pair problem (P_3, R_0, static, formative).
2.47/1.28	
2.47/1.28	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.47/1.28	
2.47/1.28	    * 0 :  
2.47/1.28	    * 1 :  
2.47/1.28	
2.47/1.28	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
2.47/1.28	
2.47/1.28	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
2.47/1.28	
2.47/1.28	We consider the dependency pair problem (P_1, R_0, static, formative).
2.47/1.28	
2.47/1.28	We apply the subterm criterion with the following projection function:
2.47/1.28	
2.47/1.28	  nu(map#) = 2 
2.47/1.28	
2.47/1.28	Thus, we can orient the dependency pairs as follows:
2.47/1.28	
2.47/1.28	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
2.47/1.28	
2.47/1.28	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.
2.47/1.28	
2.47/1.28	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
2.47/1.28	
2.47/1.28	
2.47/1.28	+++ Citations +++
2.47/1.28	
2.47/1.28	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
2.47/1.28	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
2.47/1.28	[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.
2.47/1.28	EOF