2.66/1.47	YES
2.66/1.47	We consider the system theBenchmark.
2.66/1.47	
2.66/1.47	  Alphabet:
2.66/1.47	
2.66/1.47	    0 : [] --> nat 
2.66/1.47	    cons : [nat * list] --> list 
2.66/1.47	    false : [] --> bool 
2.66/1.47	    gcd : [nat * nat] --> nat 
2.66/1.47	    gcdlists : [list * list] --> list 
2.66/1.47	    if : [bool * nat * nat] --> nat 
2.66/1.47	    le : [nat * nat] --> bool 
2.66/1.47	    minus : [nat * nat] --> nat 
2.66/1.47	    nil : [] --> list 
2.66/1.47	    s : [nat] --> nat 
2.66/1.47	    true : [] --> bool 
2.66/1.47	    zipWith : [nat -> nat -> nat * list * list] --> list 
2.66/1.47	
2.66/1.47	  Rules:
2.66/1.47	
2.66/1.47	    le(0, x) => true 
2.66/1.47	    le(s(x), 0) => false 
2.66/1.47	    le(s(x), s(y)) => le(x, y) 
2.66/1.47	    minus(x, 0) => x 
2.66/1.47	    minus(s(x), s(y)) => minus(x, y) 
2.66/1.47	    gcd(0, x) => 0 
2.66/1.47	    gcd(s(x), 0) => 0 
2.66/1.47	    gcd(s(x), s(y)) => if(le(y, x), s(x), s(y)) 
2.66/1.47	    if(true, s(x), s(y)) => gcd(minus(x, y), s(y)) 
2.66/1.47	    if(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 
2.66/1.47	    zipWith(f, x, nil) => nil 
2.66/1.47	    zipWith(f, nil, x) => nil 
2.66/1.47	    zipWith(f, cons(x, y), cons(z, u)) => cons(f x z, zipWith(f, y, u)) 
2.66/1.47	    gcdlists(x, y) => zipWith(/\z./\u.gcd(z, u), x, y) 
2.66/1.47	
2.66/1.47	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.66/1.47	
2.66/1.47	We observe that the rules contain a first-order subset:
2.66/1.47	
2.66/1.47	  le(0, X) => true 
2.66/1.47	  le(s(X), 0) => false 
2.66/1.47	  le(s(X), s(Y)) => le(X, Y) 
2.66/1.47	  minus(X, 0) => X 
2.66/1.47	  minus(s(X), s(Y)) => minus(X, Y) 
2.66/1.47	  gcd(0, X) => 0 
2.66/1.47	  gcd(s(X), 0) => 0 
2.66/1.47	  gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
2.66/1.47	  if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
2.66/1.47	  if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 
2.66/1.47	
2.66/1.47	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.
2.66/1.47	
2.66/1.47	According to the external first-order termination prover, this system is indeed Ce-terminating:
2.66/1.47	
2.66/1.47	 || proof of resources/system.trs
2.66/1.47	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.66/1.47	 || 
2.66/1.47	 || 
2.66/1.47	 || Termination w.r.t. Q of the given QTRS could be proven:
2.66/1.47	 || 
2.66/1.47	 || (0) QTRS
2.66/1.47	 || (1) DependencyPairsProof [EQUIVALENT]
2.66/1.47	 || (2) QDP
2.66/1.47	 || (3) DependencyGraphProof [EQUIVALENT]
2.66/1.47	 || (4) AND
2.66/1.47	 ||     (5) QDP
2.66/1.47	 ||         (6) UsableRulesProof [EQUIVALENT]
2.66/1.47	 ||         (7) QDP
2.66/1.47	 ||         (8) QDPSizeChangeProof [EQUIVALENT]
2.66/1.47	 ||         (9) YES
2.66/1.47	 ||     (10) QDP
2.66/1.47	 ||         (11) UsableRulesProof [EQUIVALENT]
2.66/1.47	 ||         (12) QDP
2.66/1.47	 ||         (13) QDPSizeChangeProof [EQUIVALENT]
2.66/1.47	 ||         (14) YES
2.66/1.47	 ||     (15) QDP
2.66/1.47	 ||         (16) QDPOrderProof [EQUIVALENT]
2.66/1.47	 ||         (17) QDP
2.66/1.47	 ||         (18) DependencyGraphProof [EQUIVALENT]
2.66/1.47	 ||         (19) TRUE
2.66/1.47	 || 
2.66/1.47	 || 
2.66/1.47	 || ----------------------------------------
2.66/1.47	 || 
2.66/1.47	 || (0)
2.66/1.47	 || Obligation:
2.66/1.47	 || Q restricted rewrite system:
2.66/1.47	 || The TRS R consists of the following rules:
2.66/1.47	 || 
2.66/1.47	 ||    le(0, %X) -> true
2.66/1.47	 ||    le(s(%X), 0) -> false
2.66/1.47	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.47	 ||    minus(%X, 0) -> %X
2.66/1.47	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.47	 ||    gcd(0, %X) -> 0
2.66/1.47	 ||    gcd(s(%X), 0) -> 0
2.66/1.47	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.47	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.47	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (1) DependencyPairsProof (EQUIVALENT)
2.66/1.48	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (2)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
2.66/1.48	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.66/1.48	 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    GCD(s(%X), s(%Y)) -> LE(%Y, %X)
2.66/1.48	 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    IF(true, s(%X), s(%Y)) -> MINUS(%X, %Y)
2.66/1.48	 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
2.66/1.48	 ||    IF(false, s(%X), s(%Y)) -> MINUS(%Y, %X)
2.66/1.48	 || 
2.66/1.48	 || The TRS R consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    le(0, %X) -> true
2.66/1.48	 ||    le(s(%X), 0) -> false
2.66/1.48	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 ||    gcd(0, %X) -> 0
2.66/1.48	 ||    gcd(s(%X), 0) -> 0
2.66/1.48	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (3) DependencyGraphProof (EQUIVALENT)
2.66/1.48	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (4)
2.66/1.48	 || Complex Obligation (AND)
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (5)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.66/1.48	 || 
2.66/1.48	 || The TRS R consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    le(0, %X) -> true
2.66/1.48	 ||    le(s(%X), 0) -> false
2.66/1.48	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 ||    gcd(0, %X) -> 0
2.66/1.48	 ||    gcd(s(%X), 0) -> 0
2.66/1.48	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (6) UsableRulesProof (EQUIVALENT)
2.66/1.48	 || 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.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (7)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.66/1.48	 || 
2.66/1.48	 || R is empty.
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (8) QDPSizeChangeProof (EQUIVALENT)
2.66/1.48	 || 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. 
2.66/1.48	 || 
2.66/1.48	 || From the DPs we obtained the following set of size-change graphs:
2.66/1.48	 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
2.66/1.48	 || The graph contains the following edges 1 > 1, 2 > 2
2.66/1.48	 || 
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (9)
2.66/1.48	 || YES
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (10)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
2.66/1.48	 || 
2.66/1.48	 || The TRS R consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    le(0, %X) -> true
2.66/1.48	 ||    le(s(%X), 0) -> false
2.66/1.48	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 ||    gcd(0, %X) -> 0
2.66/1.48	 ||    gcd(s(%X), 0) -> 0
2.66/1.48	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (11) UsableRulesProof (EQUIVALENT)
2.66/1.48	 || 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.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (12)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    LE(s(%X), s(%Y)) -> LE(%X, %Y)
2.66/1.48	 || 
2.66/1.48	 || R is empty.
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (13) QDPSizeChangeProof (EQUIVALENT)
2.66/1.48	 || 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. 
2.66/1.48	 || 
2.66/1.48	 || From the DPs we obtained the following set of size-change graphs:
2.66/1.48	 || *LE(s(%X), s(%Y)) -> LE(%X, %Y)
2.66/1.48	 || The graph contains the following edges 1 > 1, 2 > 2
2.66/1.48	 || 
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (14)
2.66/1.48	 || YES
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (15)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
2.66/1.48	 || 
2.66/1.48	 || The TRS R consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    le(0, %X) -> true
2.66/1.48	 ||    le(s(%X), 0) -> false
2.66/1.48	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 ||    gcd(0, %X) -> 0
2.66/1.48	 ||    gcd(s(%X), 0) -> 0
2.66/1.48	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (16) QDPOrderProof (EQUIVALENT)
2.66/1.48	 || We use the reduction pair processor [LPAR04,JAR06].
2.66/1.48	 || 
2.66/1.48	 || 
2.66/1.48	 || The following pairs can be oriented strictly and are deleted.
2.66/1.48	 || 
2.66/1.48	 ||    IF(true, s(%X), s(%Y)) -> GCD(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    IF(false, s(%X), s(%Y)) -> GCD(minus(%Y, %X), s(%X))
2.66/1.48	 || The remaining pairs can at least be oriented weakly.
2.66/1.48	 || Used ordering:  Polynomial interpretation [POLO]:
2.66/1.48	 || 
2.66/1.48	 ||    POL(0) = 0
2.66/1.48	 ||    POL(GCD(x_1, x_2)) = 1 + x_1 + x_2
2.66/1.48	 ||    POL(IF(x_1, x_2, x_3)) = 1 + x_2 + x_3
2.66/1.48	 ||    POL(false) = 0
2.66/1.48	 ||    POL(le(x_1, x_2)) = 0
2.66/1.48	 ||    POL(minus(x_1, x_2)) = x_1
2.66/1.48	 ||    POL(s(x_1)) = 1 + x_1
2.66/1.48	 ||    POL(true) = 0
2.66/1.48	 || 
2.66/1.48	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
2.66/1.48	 || 
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 || 
2.66/1.48	 || 
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (17)
2.66/1.48	 || Obligation:
2.66/1.48	 || Q DP problem:
2.66/1.48	 || The TRS P consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    GCD(s(%X), s(%Y)) -> IF(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 || 
2.66/1.48	 || The TRS R consists of the following rules:
2.66/1.48	 || 
2.66/1.48	 ||    le(0, %X) -> true
2.66/1.48	 ||    le(s(%X), 0) -> false
2.66/1.48	 ||    le(s(%X), s(%Y)) -> le(%X, %Y)
2.66/1.48	 ||    minus(%X, 0) -> %X
2.66/1.48	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
2.66/1.48	 ||    gcd(0, %X) -> 0
2.66/1.48	 ||    gcd(s(%X), 0) -> 0
2.66/1.48	 ||    gcd(s(%X), s(%Y)) -> if(le(%Y, %X), s(%X), s(%Y))
2.66/1.48	 ||    if(true, s(%X), s(%Y)) -> gcd(minus(%X, %Y), s(%Y))
2.66/1.48	 ||    if(false, s(%X), s(%Y)) -> gcd(minus(%Y, %X), s(%X))
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %X
2.66/1.48	 ||    ~PAIR(%X, %Y) -> %Y
2.66/1.48	 || 
2.66/1.48	 || Q is empty.
2.66/1.48	 || We have to consider all minimal (P,Q,R)-chains.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (18) DependencyGraphProof (EQUIVALENT)
2.66/1.48	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
2.66/1.48	 || ----------------------------------------
2.66/1.48	 || 
2.66/1.48	 || (19)
2.66/1.48	 || TRUE
2.66/1.48	 || 
2.66/1.48	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]).
2.66/1.48	
2.66/1.48	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
2.66/1.48	
2.66/1.48	  Dependency Pairs P_0:
2.66/1.48	
2.66/1.48	    0] zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   
2.66/1.48	    1] gcdlists#(X, Y) =#> zipWith#(/\x./\y.gcd(x, y), X, Y)   
2.66/1.48	    2] gcdlists#(X, Y) =#> gcd#(Z, U)   
2.66/1.48	
2.66/1.48	  Rules R_0:
2.66/1.48	
2.66/1.48	    le(0, X) => true 
2.66/1.48	    le(s(X), 0) => false 
2.66/1.48	    le(s(X), s(Y)) => le(X, Y) 
2.66/1.48	    minus(X, 0) => X 
2.66/1.48	    minus(s(X), s(Y)) => minus(X, Y) 
2.66/1.48	    gcd(0, X) => 0 
2.66/1.48	    gcd(s(X), 0) => 0 
2.66/1.48	    gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) 
2.66/1.48	    if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) 
2.66/1.48	    if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 
2.66/1.48	    zipWith(F, X, nil) => nil 
2.66/1.48	    zipWith(F, nil, X) => nil 
2.66/1.48	    zipWith(F, cons(X, Y), cons(Z, U)) => cons(F X Z, zipWith(F, Y, U)) 
2.66/1.48	    gcdlists(X, Y) => zipWith(/\x./\y.gcd(x, y), X, Y) 
2.66/1.48	
2.66/1.48	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
2.66/1.48	
2.66/1.48	We consider the dependency pair problem (P_0, R_0, static, formative).
2.66/1.48	
2.66/1.48	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
2.66/1.48	
2.66/1.48	    * 0 : 0 
2.66/1.48	    * 1 : 0 
2.66/1.48	    * 2 :  
2.66/1.48	
2.66/1.48	This graph has the following strongly connected components:
2.66/1.48	
2.66/1.48	  P_1:
2.66/1.48	
2.66/1.48	    zipWith#(F, cons(X, Y), cons(Z, U)) =#> zipWith#(F, Y, U)   
2.66/1.48	
2.66/1.48	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).
2.66/1.48	
2.66/1.48	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
2.66/1.48	
2.66/1.48	We consider the dependency pair problem (P_1, R_0, static, formative).
2.66/1.48	
2.66/1.48	We apply the subterm criterion with the following projection function:
2.66/1.48	
2.66/1.48	  nu(zipWith#) = 2 
2.66/1.48	
2.66/1.48	Thus, we can orient the dependency pairs as follows:
2.66/1.48	
2.66/1.48	  nu(zipWith#(F, cons(X, Y), cons(Z, U))) = cons(X, Y) |> Y = nu(zipWith#(F, Y, U)) 
2.66/1.48	
2.66/1.48	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.
2.66/1.48	
2.66/1.48	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
2.66/1.48	
2.66/1.48	
2.66/1.48	+++ Citations +++
2.66/1.48	
2.66/1.48	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
2.66/1.48	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
2.66/1.48	[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.
2.66/1.48	EOF