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. lexicographic path order with precedence:
    
      precedence:
      
        plus > Z > |0| > s > min > quot#
      
      argument filter:
    
        pi(quot#) = 1
        pi(s) = [1]
        pi(min) = [1]
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(Z) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        plus > |0| > s > min > Z > quot#
      
      argument filter:
    
        pi(quot#) = [1]
        pi(s) = [1]
        pi(min) = []
        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. lexicographic path order with precedence:
    
      precedence:
      
        |0| > plus > min# > s > Z > min
      
      argument filter:
    
        pi(min#) = 1
        pi(min) = 1
        pi(Z) = []
        pi(plus) = [1, 2]
        pi(s) = 1
        pi(|0|) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > plus > min# > s > Z > min
      
      argument filter:
    
        pi(min#) = [1]
        pi(min) = 1
        pi(Z) = []
        pi(plus) = [1, 2]
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

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:

  {p1}


-- Reduction pair.

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

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

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. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > plus > Z > min# > min
      
      argument filter:
    
        pi(min#) = [1]
        pi(min) = [1, 2]
        pi(Z) = []
        pi(plus) = [1, 2]
        pi(|0|) = []
        pi(s) = []
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > |0| > plus > Z > min > min#
      
      argument filter:
    
        pi(min#) = []
        pi(min) = [1, 2]
        pi(Z) = []
        pi(plus) = []
        pi(|0|) = []
        pi(s) = []
    

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: 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. lexicographic path order with precedence:
    
      precedence:
      
        s > plus#
      
      argument filter:
    
        pi(plus#) = 1
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > plus#
      
      argument filter:
    
        pi(plus#) = [1]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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