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

-- 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: plus#(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus#(minus(y,s(s(z))),minus(x,s(|0|())))
p6: plus#(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus#(plus(y,s(s(z))),plus(x,s(|0|())))

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: plus(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(|0|())))
r8: plus(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus(plus(y,s(s(z))),plus(x,s(|0|())))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}
  {p4, p5, p6}


-- 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: plus(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(|0|())))
r8: plus(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus(plus(y,s(s(z))),plus(x,s(|0|())))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      quot#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,1,1,0)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (2,1,2,1)
      minus_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,1,0)) x2 + (1,3,1,0)
      |0|_A() = (1,1,1,1)

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: 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)))
r5: plus(|0|(),y) -> y
r6: plus(s(x),y) -> s(plus(x,y))
r7: plus(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(|0|())))
r8: plus(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus(plus(y,s(s(z))),plus(x,s(|0|())))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      minus#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,0,1,1)) x2
      s_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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)
p2: plus#(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus#(plus(y,s(s(z))),plus(x,s(|0|())))
p3: plus#(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus#(minus(y,s(s(z))),minus(x,s(|0|())))

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: plus(minus(x,s(|0|())),minus(y,s(s(z)))) -> plus(minus(y,s(s(z))),minus(x,s(|0|())))
r8: plus(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus(plus(y,s(s(z))),plus(x,s(|0|())))

The set of usable rules consists of

  r1, r2, r5, r6, r7, r8

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      plus#_A(x1,x2) = x1 + ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,1,0)) x2
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + (1,0,1,2)
      plus_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + x2 + (1,1,1,1)
      |0|_A() = (0,1,1,1)
      minus_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,1,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (1,1,0,1)

The next rules are strictly ordered:

  p1, p2, p3

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