3.32/1.73	YES
3.32/1.75	We consider the system theBenchmark.
3.32/1.75	
3.32/1.75	  Alphabet:
3.32/1.75	
3.32/1.75	    0 : [] --> a 
3.32/1.75	    cons : [c * d] --> d 
3.32/1.75	    false : [] --> b 
3.32/1.75	    filter : [c -> b * d] --> d 
3.32/1.75	    filter2 : [b * c -> b * c * d] --> d 
3.32/1.75	    map : [c -> c * d] --> d 
3.32/1.75	    minus : [a * a] --> a 
3.32/1.75	    nil : [] --> d 
3.32/1.75	    plus : [a * a] --> a 
3.32/1.75	    quot : [a * a] --> a 
3.32/1.75	    s : [a] --> a 
3.32/1.75	    true : [] --> b 
3.32/1.75	
3.32/1.75	  Rules:
3.32/1.75	
3.32/1.75	    minus(x, 0) => x 
3.32/1.75	    minus(s(x), s(y)) => minus(x, y) 
3.32/1.75	    quot(0, s(x)) => 0 
3.32/1.75	    quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) 
3.32/1.75	    plus(0, x) => x 
3.32/1.75	    plus(s(x), y) => s(plus(x, y)) 
3.32/1.75	    plus(minus(x, s(0)), minus(y, s(s(z)))) => plus(minus(y, s(s(z))), minus(x, s(0))) 
3.32/1.75	    plus(plus(x, s(0)), plus(y, s(s(z)))) => plus(plus(y, s(s(z))), plus(x, s(0))) 
3.32/1.75	    map(f, nil) => nil 
3.32/1.75	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
3.32/1.75	    filter(f, nil) => nil 
3.32/1.75	    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
3.32/1.75	    filter2(true, f, x, y) => cons(x, filter(f, y)) 
3.32/1.75	    filter2(false, f, x, y) => filter(f, y) 
3.32/1.75	
3.32/1.75	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
3.32/1.75	
3.32/1.75	We observe that the rules contain a first-order subset:
3.32/1.75	
3.32/1.75	  minus(X, 0) => X 
3.32/1.75	  minus(s(X), s(Y)) => minus(X, Y) 
3.32/1.75	  quot(0, s(X)) => 0 
3.32/1.75	  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
3.32/1.75	  plus(0, X) => X 
3.32/1.75	  plus(s(X), Y) => s(plus(X, Y)) 
3.32/1.75	  plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) 
3.32/1.75	  plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 
3.32/1.75	
3.32/1.75	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.
3.32/1.75	
3.32/1.75	According to the external first-order termination prover, this system is indeed Ce-terminating:
3.32/1.75	
3.32/1.75	 || proof of resources/system.trs
3.32/1.75	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || Termination w.r.t. Q of the given QTRS could be proven:
3.32/1.75	 || 
3.32/1.75	 || (0) QTRS
3.32/1.75	 || (1) DependencyPairsProof [EQUIVALENT]
3.32/1.75	 || (2) QDP
3.32/1.75	 || (3) DependencyGraphProof [EQUIVALENT]
3.32/1.75	 || (4) AND
3.32/1.75	 ||     (5) QDP
3.32/1.75	 ||         (6) UsableRulesProof [EQUIVALENT]
3.32/1.75	 ||         (7) QDP
3.32/1.75	 ||         (8) MRRProof [EQUIVALENT]
3.32/1.75	 ||         (9) QDP
3.32/1.75	 ||         (10) MRRProof [EQUIVALENT]
3.32/1.75	 ||         (11) QDP
3.32/1.75	 ||         (12) MRRProof [EQUIVALENT]
3.32/1.75	 ||         (13) QDP
3.32/1.75	 ||         (14) DependencyGraphProof [EQUIVALENT]
3.32/1.75	 ||         (15) TRUE
3.32/1.75	 ||     (16) QDP
3.32/1.75	 ||         (17) UsableRulesProof [EQUIVALENT]
3.32/1.75	 ||         (18) QDP
3.32/1.75	 ||         (19) QDPSizeChangeProof [EQUIVALENT]
3.32/1.75	 ||         (20) YES
3.32/1.75	 ||     (21) QDP
3.32/1.75	 ||         (22) QDPOrderProof [EQUIVALENT]
3.32/1.75	 ||         (23) QDP
3.32/1.75	 ||         (24) PisEmptyProof [EQUIVALENT]
3.32/1.75	 ||         (25) YES
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (0)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q restricted rewrite system:
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (1) DependencyPairsProof (EQUIVALENT)
3.32/1.75	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (2)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
3.32/1.75	 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
3.32/1.75	 ||    QUOT(s(%X), s(%Y)) -> MINUS(%X, %Y)
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (3) DependencyGraphProof (EQUIVALENT)
3.32/1.75	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 1 less node.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (4)
3.32/1.75	 || Complex Obligation (AND)
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (5)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (6) UsableRulesProof (EQUIVALENT)
3.32/1.75	 || 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.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (7)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (8) MRRProof (EQUIVALENT)
3.32/1.75	 || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || Strictly oriented rules of the TRS R:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 || 
3.32/1.75	 || Used ordering: Polynomial interpretation [POLO]:
3.32/1.75	 || 
3.32/1.75	 ||    POL(0) = 0
3.32/1.75	 ||    POL(PLUS(x_1, x_2)) = 2*x_1 + 2*x_2
3.32/1.75	 ||    POL(minus(x_1, x_2)) = 2 + x_1 + x_2
3.32/1.75	 ||    POL(plus(x_1, x_2)) = x_1 + x_2
3.32/1.75	 ||    POL(s(x_1)) = x_1
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (9)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (10) MRRProof (EQUIVALENT)
3.32/1.75	 || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || Strictly oriented rules of the TRS R:
3.32/1.75	 || 
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 || 
3.32/1.75	 || Used ordering: Polynomial interpretation [POLO]:
3.32/1.75	 || 
3.32/1.75	 ||    POL(0) = 0
3.32/1.75	 ||    POL(PLUS(x_1, x_2)) = x_1 + x_2
3.32/1.75	 ||    POL(minus(x_1, x_2)) = 2*x_1 + 2*x_2
3.32/1.75	 ||    POL(plus(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
3.32/1.75	 ||    POL(s(x_1)) = x_1
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (11)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (12) MRRProof (EQUIVALENT)
3.32/1.75	 || By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
3.32/1.75	 || 
3.32/1.75	 || Strictly oriented dependency pairs:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(s(%X), %Y) -> PLUS(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || Strictly oriented rules of the TRS R:
3.32/1.75	 || 
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || Used ordering: Polynomial interpretation [POLO]:
3.32/1.75	 || 
3.32/1.75	 ||    POL(0) = 0
3.32/1.75	 ||    POL(PLUS(x_1, x_2)) = 2*x_1 + 2*x_2
3.32/1.75	 ||    POL(minus(x_1, x_2)) = 2*x_1 + 2*x_2
3.32/1.75	 ||    POL(plus(x_1, x_2)) = x_1 + x_2
3.32/1.75	 ||    POL(s(x_1)) = 1 + x_1
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (13)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    PLUS(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> PLUS(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    PLUS(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> PLUS(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (14) DependencyGraphProof (EQUIVALENT)
3.32/1.75	 || The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (15)
3.32/1.75	 || TRUE
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (16)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (17) UsableRulesProof (EQUIVALENT)
3.32/1.75	 || 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.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (18)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || R is empty.
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (19) QDPSizeChangeProof (EQUIVALENT)
3.32/1.75	 || 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. 
3.32/1.75	 || 
3.32/1.75	 || From the DPs we obtained the following set of size-change graphs:
3.32/1.75	 || *MINUS(s(%X), s(%Y)) -> MINUS(%X, %Y)
3.32/1.75	 || The graph contains the following edges 1 > 1, 2 > 2
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (20)
3.32/1.75	 || YES
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (21)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || The TRS P consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
3.32/1.75	 || 
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (22) QDPOrderProof (EQUIVALENT)
3.32/1.75	 || We use the reduction pair processor [LPAR04,JAR06].
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || The following pairs can be oriented strictly and are deleted.
3.32/1.75	 || 
3.32/1.75	 ||    QUOT(s(%X), s(%Y)) -> QUOT(minus(%X, %Y), s(%Y))
3.32/1.75	 || The remaining pairs can at least be oriented weakly.
3.32/1.75	 || Used ordering:  Polynomial interpretation [POLO]:
3.32/1.75	 || 
3.32/1.75	 ||    POL(0) = 0
3.32/1.75	 ||    POL(QUOT(x_1, x_2)) = x_1
3.32/1.75	 ||    POL(minus(x_1, x_2)) = x_1
3.32/1.75	 ||    POL(s(x_1)) = 1 + x_1
3.32/1.75	 || 
3.32/1.75	 || The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 || 
3.32/1.75	 || 
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (23)
3.32/1.75	 || Obligation:
3.32/1.75	 || Q DP problem:
3.32/1.75	 || P is empty.
3.32/1.75	 || The TRS R consists of the following rules:
3.32/1.75	 || 
3.32/1.75	 ||    minus(%X, 0) -> %X
3.32/1.75	 ||    minus(s(%X), s(%Y)) -> minus(%X, %Y)
3.32/1.75	 ||    quot(0, s(%X)) -> 0
3.32/1.75	 ||    quot(s(%X), s(%Y)) -> s(quot(minus(%X, %Y), s(%Y)))
3.32/1.75	 ||    plus(0, %X) -> %X
3.32/1.75	 ||    plus(s(%X), %Y) -> s(plus(%X, %Y))
3.32/1.75	 ||    plus(minus(%X, s(0)), minus(%Y, s(s(%Z)))) -> plus(minus(%Y, s(s(%Z))), minus(%X, s(0)))
3.32/1.75	 ||    plus(plus(%X, s(0)), plus(%Y, s(s(%Z)))) -> plus(plus(%Y, s(s(%Z))), plus(%X, s(0)))
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %X
3.32/1.75	 ||    ~PAIR(%X, %Y) -> %Y
3.32/1.75	 || 
3.32/1.75	 || Q is empty.
3.32/1.75	 || We have to consider all minimal (P,Q,R)-chains.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (24) PisEmptyProof (EQUIVALENT)
3.32/1.75	 || The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.32/1.75	 || ----------------------------------------
3.32/1.75	 || 
3.32/1.75	 || (25)
3.32/1.75	 || YES
3.32/1.75	 || 
3.32/1.75	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]).
3.32/1.75	
3.32/1.75	We thus obtain the following dependency pair problem (P_0, R_0, static, formative):
3.32/1.75	
3.32/1.75	  Dependency Pairs P_0:
3.32/1.75	
3.32/1.75	    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
3.32/1.75	    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
3.32/1.75	    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	
3.32/1.75	  Rules R_0:
3.32/1.75	
3.32/1.75	    minus(X, 0) => X 
3.32/1.75	    minus(s(X), s(Y)) => minus(X, Y) 
3.32/1.75	    quot(0, s(X)) => 0 
3.32/1.75	    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
3.32/1.75	    plus(0, X) => X 
3.32/1.75	    plus(s(X), Y) => s(plus(X, Y)) 
3.32/1.75	    plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) 
3.32/1.75	    plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 
3.32/1.75	    map(F, nil) => nil 
3.32/1.75	    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
3.32/1.75	    filter(F, nil) => nil 
3.32/1.75	    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
3.32/1.75	    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
3.32/1.75	    filter2(false, F, X, Y) => filter(F, Y) 
3.32/1.75	
3.32/1.75	Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.
3.32/1.75	
3.32/1.75	We consider the dependency pair problem (P_0, R_0, static, formative).
3.32/1.75	
3.32/1.75	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
3.32/1.75	
3.32/1.75	    * 0 : 0 
3.32/1.75	    * 1 : 2, 3 
3.32/1.75	    * 2 : 1 
3.32/1.75	    * 3 : 1 
3.32/1.75	
3.32/1.75	This graph has the following strongly connected components:
3.32/1.75	
3.32/1.75	  P_1:
3.32/1.75	
3.32/1.75	    map#(F, cons(X, Y)) =#> map#(F, Y)   
3.32/1.75	
3.32/1.75	  P_2:
3.32/1.75	
3.32/1.75	    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
3.32/1.75	    filter2#(true, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	    filter2#(false, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	
3.32/1.75	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).
3.32/1.75	
3.32/1.75	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.
3.32/1.75	
3.32/1.75	We consider the dependency pair problem (P_2, R_0, static, formative).
3.32/1.75	
3.32/1.75	We apply the subterm criterion with the following projection function:
3.32/1.75	
3.32/1.75	  nu(filter2#) = 4 
3.32/1.75	  nu(filter#) = 2 
3.32/1.75	
3.32/1.75	Thus, we can orient the dependency pairs as follows:
3.32/1.75	
3.32/1.75	  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
3.32/1.75	  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
3.32/1.75	  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
3.32/1.75	
3.32/1.75	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:
3.32/1.75	
3.32/1.75	  filter2#(true, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	  filter2#(false, F, X, Y) =#> filter#(F, Y)   
3.32/1.75	
3.32/1.75	Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_3, R_0, static, formative) is finite.
3.32/1.75	
3.32/1.75	We consider the dependency pair problem (P_3, R_0, static, formative).
3.32/1.75	
3.32/1.75	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
3.32/1.75	
3.32/1.75	    * 0 :  
3.32/1.75	    * 1 :  
3.32/1.75	
3.32/1.75	This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.
3.32/1.75	
3.32/1.75	Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.
3.32/1.75	
3.32/1.75	We consider the dependency pair problem (P_1, R_0, static, formative).
3.32/1.75	
3.32/1.75	We apply the subterm criterion with the following projection function:
3.32/1.75	
3.32/1.75	  nu(map#) = 2 
3.32/1.75	
3.32/1.75	Thus, we can orient the dependency pairs as follows:
3.32/1.75	
3.32/1.75	  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 
3.32/1.75	
3.32/1.75	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.
3.32/1.75	
3.32/1.75	As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
3.32/1.75	
3.32/1.75	
3.32/1.75	+++ Citations +++
3.32/1.75	
3.32/1.75	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
3.32/1.75	[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
3.32/1.75	[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.
3.32/1.75	EOF