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, r3, 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) = x2 + 1
          s_A(x1) = x1
          minus_A(x1,x2) = x1 + 3
          |0|_A() = 1
          quot_A(x1,x2) = 2
          plus_A(x1,x2) = x2 + 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          quot#_A(x1,x2) = max{0, x1 - 2, x2 - 16}
          s_A(x1) = x1 + 14
          minus_A(x1,x2) = x1 + 4
          |0|_A = 0
          quot_A(x1,x2) = max{7, x1 - 16, x2 - 29}
          plus_A(x1,x2) = max{4, x1 - 21, x2 - 21}
          quot#!_A(x1,x2) = max{1, x1 - 5}
          s!_A(x1) = x1 + 20
          minus!_A(x1,x2) = max{21, x1 - 13}
          |0|!_A = 21
          quot!_A(x1,x2) = max{x1 + 4, x2 - 15}
          plus!_A(x1,x2) = max{25, x2 - 2}
  
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = []
      pi(|0|) = []
      pi(quot) = []
      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 + 7
          plus_A(x1,x2) = x1 + x2 + 3
          |0|_A() = 1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = max{0, x1 - 9, x2 - 9}
          s_A(x1) = max{3, x1}
          minus_A(x1,x2) = max{x1 + 7, x2 - 9}
          plus_A(x1,x2) = max{8, x1 - 9, x2}
          |0|_A = 2
          minus#!_A(x1,x2) = max{5, x1 - 9, x2 - 9}
          s!_A(x1) = max{2, x1 - 30}
          minus!_A(x1,x2) = max{4, x1 - 10, x2 - 28}
          plus!_A(x1,x2) = max{3, x1 - 26, x2 - 29}
          |0|!_A = 0
  
    partial status:
  
      pi(minus#) = [1]
      pi(s) = [1]
      pi(minus) = [1, 2]
      pi(plus) = [1]
      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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          plus#_A(x1,x2) = x1
          s_A(x1) = x1 + 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          plus#_A(x1,x2) = 0
          s_A(x1) = max{5, x1 - 2}
          plus#!_A(x1,x2) = max{2, x1 - 32, x2 - 1}
          s!_A(x1) = max{1, x1 - 32}
  
    partial status:
  
      pi(plus#) = [1]
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.