YES
We show the termination of the TRS R:

  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  quot(|0|(),s(y)) -> |0|()
  quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))


-- SCC decomposition.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)
p2: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))
p3: quot#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y))

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        quot#_A(x1,x2) = max{5, x1 + 4, x2 - 3}
        s_A(x1) = x1 + 9
        minus_A(x1,x2) = max{7, x1 + 6}
        |0|_A = 3
        quot_A(x1,x2) = max{8, x2 - 14}
  
    base order 2:
  
      precedence:
      
        quot# > s = minus > quot > |0|
      
  
    partial status:
  
      pi(quot#) = [1]
      pi(s) = [1]
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

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

p1: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: minus(x,|0|()) -> x
r2: minus(s(x),s(y)) -> minus(x,y)
r3: quot(|0|(),s(y)) -> |0|()
r4: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))

The set of usable rules consists of

  r3

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{5, x1 + 2}
        s_A(x1) = max{6, x1 + 4}
        quot_A(x1,x2) = max{7, x1 - 13, x2 - 11}
        |0|_A = 5
  
    base order 2:
  
      precedence:
      
        quot > minus# = s > |0|
      
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(quot) = []
      pi(|0|) = []

The following rules are removed:

  p1

Then no dependency pair remains.