YES

We show the termination of the TRS R:

  plus(|0|(),Y) -> Y
  plus(s(X),Y) -> s(plus(X,Y))
  min(X,|0|()) -> X
  min(s(X),s(Y)) -> min(X,Y)
  min(min(X,Y),Z()) -> min(X,plus(Y,Z()))
  quot(|0|(),s(Y)) -> |0|()
  quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(X),Y) -> plus#(X,Y)
p2: min#(s(X),s(Y)) -> min#(X,Y)
p3: min#(min(X,Y),Z()) -> min#(X,plus(Y,Z()))
p4: min#(min(X,Y),Z()) -> plus#(Y,Z())
p5: quot#(s(X),s(Y)) -> quot#(min(X,Y),s(Y))
p6: quot#(s(X),s(Y)) -> min#(X,Y)

and R consists of:

r1: plus(|0|(),Y) -> Y
r2: plus(s(X),Y) -> s(plus(X,Y))
r3: min(X,|0|()) -> X
r4: min(s(X),s(Y)) -> min(X,Y)
r5: min(min(X,Y),Z()) -> min(X,plus(Y,Z()))
r6: quot(|0|(),s(Y)) -> |0|()
r7: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

The estimated dependency graph contains the following SCCs:

  {p5}
  {p2, p3}
  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(s(X),s(Y)) -> quot#(min(X,Y),s(Y))

and R consists of:

r1: plus(|0|(),Y) -> Y
r2: plus(s(X),Y) -> s(plus(X,Y))
r3: min(X,|0|()) -> X
r4: min(s(X),s(Y)) -> min(X,Y)
r5: min(min(X,Y),Z()) -> min(X,plus(Y,Z()))
r6: quot(|0|(),s(Y)) -> |0|()
r7: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            quot#_A(x1,x2) = x1
            s_A(x1) = x1 + (3,2)
            min_A(x1,x2) = x1 + (2,2)
            plus_A(x1,x2) = ((1,1),(1,0)) x1 + x2 + (4,4)
            |0|_A() = (1,1)
            Z_A() = (3,1)
    
      precedence:
      
        s > quot# = min = plus = |0| = Z
      
      partial status:
    
        pi(quot#) = [1]
        pi(s) = [1]
        pi(min) = [1]
        pi(plus) = [2]
        pi(|0|) = []
        pi(Z) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            quot#_A(x1,x2) = (0,0)
            s_A(x1) = ((0,1),(0,0)) x1 + (1,1)
            min_A(x1,x2) = x1 + (4,1)
            plus_A(x1,x2) = (3,1)
            |0|_A() = (0,1)
            Z_A() = (0,0)
    
      precedence:
      
        plus > Z > s = min > |0| > quot#
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(min) = [1]
        pi(plus) = []
        pi(|0|) = []
        pi(Z) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: min#(min(X,Y),Z()) -> min#(X,plus(Y,Z()))
p2: min#(s(X),s(Y)) -> min#(X,Y)

and R consists of:

r1: plus(|0|(),Y) -> Y
r2: plus(s(X),Y) -> s(plus(X,Y))
r3: min(X,|0|()) -> X
r4: min(s(X),s(Y)) -> min(X,Y)
r5: min(min(X,Y),Z()) -> min(X,plus(Y,Z()))
r6: quot(|0|(),s(Y)) -> |0|()
r7: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1,x2) = max{x1, x2 + 5}
          min_A(x1,x2) = max{x1 + 9, x2 + 9}
          Z_A = 0
          plus_A(x1,x2) = max{x1 + 3, x2 + 1}
          s_A(x1) = x1
          |0|_A = 0
    
      precedence:
      
        min# = plus > s = |0| > min = Z
      
      partial status:
    
        pi(min#) = [1, 2]
        pi(min) = [1, 2]
        pi(Z) = []
        pi(plus) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          min#_A(x1,x2) = max{1, x1 - 3, x2 - 2}
          min_A(x1,x2) = max{x1 + 13, x2}
          Z_A = 9
          plus_A(x1,x2) = max{8, x1 - 2, x2 - 13}
          s_A(x1) = max{9, x1 + 1}
          |0|_A = 0
    
      precedence:
      
        min# = min = Z = plus = s = |0|
      
      partial status:
    
        pi(min#) = []
        pi(min) = [2]
        pi(Z) = []
        pi(plus) = []
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: plus#(s(X),Y) -> plus#(X,Y)

and R consists of:

r1: plus(|0|(),Y) -> Y
r2: plus(s(X),Y) -> s(plus(X,Y))
r3: min(X,|0|()) -> X
r4: min(s(X),s(Y)) -> min(X,Y)
r5: min(min(X,Y),Z()) -> min(X,plus(Y,Z()))
r6: quot(|0|(),s(Y)) -> |0|()
r7: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{2, x1 + 1, x2}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          plus#_A(x1,x2) = max{x1 - 1, x2 + 1}
          s_A(x1) = x1
    
      precedence:
      
        plus# = s
      
      partial status:
    
        pi(plus#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.