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:

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

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:

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

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

The next rules are strictly ordered:

  p1

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