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, r7

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          quot#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 2
          minus_A(x1,x2) = x1 + 1
          |0|_A() = 1
          plus_A(x1,x2) = x2 + 3
  
    base order 2:
  
      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 + 2
            minus_A(x1,x2) = x1 + 1
            |0|_A() = 1
            plus_A(x1,x2) = x2 + 2
            quot#!_A(x1,x2) = x2
            s!_A(x1) = x1 + 4
            minus!_A(x1,x2) = x1 + 3
            |0|!_A() = 0
            plus!_A(x1,x2) = 5
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(plus) = []

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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          minus#_A(x1,x2) = x1 + x2 + 3
          s_A(x1) = x1 + 2
          minus_A(x1,x2) = x1 + x2 + 4
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 1
  
    base order 2:
  
      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) = x1 + x2
            s_A(x1) = x1 + 1
            minus_A(x1,x2) = x1 + x2 + 6
            plus_A(x1,x2) = x1 + x2 + 1
            |0|_A() = 1
            minus#!_A(x1,x2) = x1 + x2 + 5
            s!_A(x1) = x1 + 6
            minus!_A(x1,x2) = x1 + x2 + 3
            plus!_A(x1,x2) = x1 + x2 + 10
            |0|!_A() = 3
  
    partial status:
  
      pi(minus#) = []
      pi(s) = [1]
      pi(minus) = [1, 2]
      pi(plus) = []
      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

  r1, r3

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1 + x2 + 2
          s_A(x1) = x1 + 1
          minus_A(x1,x2) = x1 + 2
          |0|_A() = 0
          quot_A(x1,x2) = x2
  
    base order 2:
  
      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
            minus_A(x1,x2) = x1 + x2 + 2
            |0|_A() = 1
            quot_A(x1,x2) = x1
            plus#!_A(x1,x2) = x1 + x2
            s!_A(x1) = x1 + 2
            minus!_A(x1,x2) = x1 + x2 + 3
            |0|!_A() = 3
            quot!_A(x1,x2) = x2
  
    partial status:
  
      pi(plus#) = [2]
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []

The following rules are removed:

  p1

Then no dependency pair remains.