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)))
  plus(|0|(),y) -> y
  plus(s(x),y) -> s(plus(x,y))
  minus(minus(x,y),z) -> minus(x,plus(y,z))


-- 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)
p4: plus#(s(x),y) -> plus#(x,y)
p5: minus#(minus(x,y),z) -> minus#(x,plus(y,z))
p6: minus#(minus(x,y),z) -> plus#(y,z)

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)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}
  {p4}


-- 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)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))

The set of usable rules consists of

  r1, r2, r5, r6, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              quot#_A(x1,x2) = x1
              s_A(x1) = x1 + 3
              minus_A(x1,x2) = x1 + 1
              |0|_A() = 2
              plus_A(x1,x2) = x1 + x2 + 1
              quot#!_A(x1,x2) = x1 + 1
              s!_A(x1) = x1 + 3
              minus!_A(x1,x2) = x1 + 5
              |0|!_A() = 0
              plus!_A(x1,x2) = x1 + x2 + 4
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            quot#_A(x1,x2) = max{0, x1 - 7, x2 - 6}
            s_A(x1) = max{4, x1}
            minus_A(x1,x2) = max{6, x1, x2 - 8}
            |0|_A = 4
            plus_A(x1,x2) = max{6, x1 - 5, x2}
            quot#!_A(x1,x2) = 0
            s!_A(x1) = max{12, x1 - 8}
            minus!_A(x1,x2) = max{14, x1 - 7, x2 - 22}
            |0|!_A = 3
            plus!_A(x1,x2) = max{13, x1 - 7, x2 - 3}
        
  
    partial status:
  
      pi(quot#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []
      pi(plus) = [1, 2]

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)
p2: minus#(minus(x,y),z) -> minus#(x,plus(y,z))

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)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))

The set of usable rules consists of

  r1, r5, r6

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              minus#_A(x1,x2) = 0
              s_A(x1) = 1
              minus_A(x1,x2) = x1 + x2 + 2
              plus_A(x1,x2) = x1 + x2 + 3
              |0|_A() = 1
              minus#!_A(x1,x2) = 1
              s!_A(x1) = 2
              minus!_A(x1,x2) = 0
              plus!_A(x1,x2) = 3
              |0|!_A() = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            minus#_A(x1,x2) = max{0, x1 - 10, x2 - 10}
            s_A(x1) = max{2, x1}
            minus_A(x1,x2) = max{x1 + 8, x2 - 10}
            plus_A(x1,x2) = max{9, x1 - 10, x2}
            |0|_A = 11
            minus#!_A(x1,x2) = max{3, x1 - 3, x2 - 23}
            s!_A(x1) = max{1, x1 - 10}
            minus!_A(x1,x2) = max{2, x1 - 13, x2 - 34}
            plus!_A(x1,x2) = max{4, x1 - 9, x2 - 24}
            |0|!_A = 0
        
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(minus) = [1]
      pi(plus) = [1, 2]
      pi(|0|) = []

The following rules are removed:

  p1, p2

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: 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)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: minus(minus(x,y),z) -> minus(x,plus(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              plus#_A(x1,x2) = x1
              s_A(x1) = x1 + 1
              plus#!_A(x1,x2) = 0
              s!_A(x1) = 2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            plus#_A(x1,x2) = 0
            s_A(x1) = max{17, x1 - 5}
            plus#!_A(x1,x2) = x2 + 9
            s!_A(x1) = max{7, x1 + 3}
        
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.