We consider the system Applicative_first_order_05__#3.40.

  Alphabet:

    0 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    minus : [a * a] --> a 
    nil : [] --> d 
    plus : [a * a] --> a 
    quot : [a * a] --> a 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    quot(0, s(x)) => 0 
    quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) 
    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    plus(minus(x, s(0)), minus(y, s(s(z)))) => plus(minus(y, s(s(z))), minus(x, s(0))) 
    plus(plus(x, s(0)), plus(y, s(s(z)))) => plus(plus(y, s(s(z))), plus(x, s(0))) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(f, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We observe that the rules contain a first-order subset:

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  quot(0, s(X)) => 0 
  quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) 
  plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 

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.

According to mutermprover, this system is indeed Ce-terminating:

 || 
 || Problem 1: 
 || 
 || (VAR %X %Y %Z)
 || (RULES
 || minus(s(%X),s(%Y)) -> minus(%X,%Y)
 || minus(%X,0) -> %X
 || plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 || plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 || plus(0,%X) -> %X
 || plus(s(%X),%Y) -> s(plus(%X,%Y))
 || quot(0,s(%X)) -> 0
 || quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 || ~PAIR(%X,%Y) -> %X
 || ~PAIR(%X,%Y) -> %Y
 || )
 || 
 || Problem 1: 
 || 
 || Dependency Pairs Processor:
 || -> Pairs:
 ||  MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y)
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  PLUS(s(%X),%Y) -> PLUS(%X,%Y)
 ||  QUOT(s(%X),s(%Y)) -> MINUS(%X,%Y)
 ||  QUOT(s(%X),s(%Y)) -> QUOT(minus(%X,%Y),s(%Y))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y)
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  PLUS(s(%X),%Y) -> PLUS(%X,%Y)
 ||  QUOT(s(%X),s(%Y)) -> MINUS(%X,%Y)
 ||  QUOT(s(%X),s(%Y)) -> QUOT(minus(%X,%Y),s(%Y))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  PLUS(s(%X),%Y) -> PLUS(%X,%Y)
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->->Cycle:
 || ->->-> Pairs:
 ||  MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y)
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->->Cycle:
 || ->->-> Pairs:
 ||  QUOT(s(%X),s(%Y)) -> QUOT(minus(%X,%Y),s(%Y))
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || 
 || The problem is decomposed in 3 subproblems.
 || 
 || Problem 1.1: 
 || 
 || Reduction Pair Processor:
 || -> Pairs:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  PLUS(s(%X),%Y) -> PLUS(%X,%Y)
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || -> Usable rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [minus](X1,X2) = X1 + 2.X2
 || [plus](X1,X2) = 2.X1 + 2.X2 + 2
 || [0] = 2
 || [s](X) = X + 2
 || [PLUS](X1,X2) = X1 + X2
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1.1: 
 || 
 || Narrowing Processor:
 || -> Pairs:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Narrowed Pairs:
 || ->->Original Pair:
 ||  PLUS(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> PLUS(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 || ->-> Narrowed pairs:
 ||  PLUS(minus(s(%X),s(0)),minus(x4,s(s(x5)))) -> PLUS(minus(x4,s(s(x5))),minus(%X,0))
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || ->->Original Pair:
 ||  PLUS(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> PLUS(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 || ->-> Narrowed pairs:
 ||  PLUS(plus(0,s(0)),plus(x7,s(s(x8)))) -> PLUS(plus(x7,s(s(x8))),s(0))
 ||  PLUS(plus(s(%X),s(0)),plus(x7,s(s(x8)))) -> PLUS(plus(x7,s(s(x8))),s(plus(%X,s(0))))
 ||  PLUS(plus(x6,s(0)),plus(0,s(s(x8)))) -> PLUS(s(s(x8)),plus(x6,s(0)))
 ||  PLUS(plus(x6,s(0)),plus(s(%X),s(s(x8)))) -> PLUS(s(plus(%X,s(s(x8)))),plus(x6,s(0)))
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  PLUS(minus(s(%X),s(0)),minus(x4,s(s(x5)))) -> PLUS(minus(x4,s(s(x5))),minus(%X,0))
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 ||  PLUS(plus(0,s(0)),plus(x7,s(s(x8)))) -> PLUS(plus(x7,s(s(x8))),s(0))
 ||  PLUS(plus(s(%X),s(0)),plus(x7,s(s(x8)))) -> PLUS(plus(x7,s(s(x8))),s(plus(%X,s(0))))
 ||  PLUS(plus(x6,s(0)),plus(0,s(s(x8)))) -> PLUS(s(s(x8)),plus(x6,s(0)))
 ||  PLUS(plus(x6,s(0)),plus(s(%X),s(s(x8)))) -> PLUS(s(plus(%X,s(s(x8)))),plus(x6,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  PLUS(minus(s(%X),s(0)),minus(x4,s(s(x5)))) -> PLUS(minus(x4,s(s(x5))),minus(%X,0))
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1.1: 
 || 
 || Reduction Pair Processor:
 || -> Pairs:
 ||  PLUS(minus(s(%X),s(0)),minus(x4,s(s(x5)))) -> PLUS(minus(x4,s(s(x5))),minus(%X,0))
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || -> Usable rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [minus](X1,X2) = 2.X1 + X2
 || [0] = 2
 || [s](X) = 2.X
 || [PLUS](X1,X2) = 2.X1 + 2.X2
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 || ->->Cycle:
 || ->->-> Pairs:
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || ->->-> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || 
 || Problem 1.1: 
 || 
 || Reduction Pair Processor:
 || -> Pairs:
 ||  PLUS(minus(x3,s(0)),minus(s(%X),s(s(x5)))) -> PLUS(minus(%X,s(x5)),minus(x3,s(0)))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || -> Usable rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [minus](X1,X2) = 2.X1
 || [0] = 0
 || [s](X) = X + 2
 || [PLUS](X1,X2) = X1 + X2
 || 
 || Problem 1.1: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.2: 
 || 
 || Subterm Processor:
 || -> Pairs:
 ||  MINUS(s(%X),s(%Y)) -> MINUS(%X,%Y)
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Projection:
 ||  pi(MINUS) = 1
 || 
 || Problem 1.2: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
 || Problem 1.3: 
 || 
 || Reduction Pair Processor:
 || -> Pairs:
 ||  QUOT(s(%X),s(%Y)) -> QUOT(minus(%X,%Y),s(%Y))
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || -> Usable rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 || ->Interpretation type:
 ||  Linear
 || ->Coefficients:
 ||  Natural Numbers
 || ->Dimension:
 ||  1
 || ->Bound:
 ||  2
 || ->Interpretation:
 ||  
 || [minus](X1,X2) = 2.X1 + 1
 || [0] = 0
 || [s](X) = 2.X + 2
 || [QUOT](X1,X2) = 2.X1
 || 
 || Problem 1.3: 
 || 
 || SCC Processor:
 || -> Pairs:
 ||  Empty
 || -> Rules:
 ||  minus(s(%X),s(%Y)) -> minus(%X,%Y)
 ||  minus(%X,0) -> %X
 ||  plus(minus(%X,s(0)),minus(%Y,s(s(%Z)))) -> plus(minus(%Y,s(s(%Z))),minus(%X,s(0)))
 ||  plus(plus(%X,s(0)),plus(%Y,s(s(%Z)))) -> plus(plus(%Y,s(s(%Z))),plus(%X,s(0)))
 ||  plus(0,%X) -> %X
 ||  plus(s(%X),%Y) -> s(plus(%X,%Y))
 ||  quot(0,s(%X)) -> 0
 ||  quot(s(%X),s(%Y)) -> s(quot(minus(%X,%Y),s(%Y)))
 ||  ~PAIR(%X,%Y) -> %X
 ||  ~PAIR(%X,%Y) -> %Y
 || ->Strongly Connected Components:
 ||  There is no strongly connected component
 || 
 || The problem is finite.
 || 
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 [FuhKop19]).

We thus obtain the following dependency pair problem (P_0, R_0, computable, formative):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
    1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    2] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    3] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    quot(0, s(X)) => 0 
    quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) 
    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) 
    plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(F, Y) 

Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_0, R_0, computable, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 : 2, 3 
    * 2 : 1 
    * 3 : 1 

This graph has the following strongly connected components:

  P_1:

    map#(F, cons(X, Y)) =#> map#(F, Y)   

  P_2:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(F, Y)   

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).

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_2, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by (P_3, R_0, computable, f), where P_3 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_3, R_0, computable, formative).

We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

Thus, the original system is terminating if (P_1, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, computable, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[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.