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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = x1 + (2,1)
        min_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
        plus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,0)
        |0|_A() = (1,1)
        Z_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((0,0),(1,0)) x1
        s_A(x1) = (1,1)
        min_A(x1,x2) = ((1,1),(1,0)) x1 + (2,1)
        plus_A(x1,x2) = x1 + (1,1)
        |0|_A() = (1,1)
        Z_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = (0,0)
        s_A(x1) = (1,1)
        min_A(x1,x2) = x1 + (1,1)
        plus_A(x1,x2) = ((0,0),(1,0)) x1 + (2,1)
        |0|_A() = (1,1)
        Z_A() = (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2
        min_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (4,1)
        Z_A() = (1,1)
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        s_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = x1 + ((1,0),(1,0)) x2
        min_A(x1,x2) = x1 + (2,0)
        Z_A() = (1,1)
        plus_A(x1,x2) = x1 + (2,1)
        s_A(x1) = (1,2)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        min#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2
        min_A(x1,x2) = x1 + (1,1)
        Z_A() = (1,1)
        plus_A(x1,x2) = ((0,1),(0,0)) x1 + (3,1)
        s_A(x1) = (1,2)
        |0|_A() = (1,1)
    

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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,1)) x1
        s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,0),(1,0)) x1
        s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

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