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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,0),(1,0)) x1
        s_A(x1) = ((1,1),(0,1)) x1 + (2,1)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = x1
        s_A(x1) = x1 + (2,1)
        minus_A(x1,x2) = ((1,1),(0,1)) x1 + (1,2)
        |0|_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        quot#_A(x1,x2) = ((1,1),(1,1)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
        |0|_A() = (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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        minus#_A(x1,x2) = (0,0)
        s_A(x1) = ((0,1),(0,0)) x1 + (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: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + (1,1)
        plus_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (0,1)
        |0|_A() = (1,1)
        minus_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,0),(1,1)) x2 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,1),(1,0)) x2
        s_A(x1) = ((0,0),(1,0)) x1 + (1,1)
        plus_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (0,1)
        |0|_A() = (1,1)
        minus_A(x1,x2) = ((1,1),(0,0)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,1),(0,0)) x1
        s_A(x1) = (1,1)
        plus_A(x1,x2) = (2,1)
        |0|_A() = (0,1)
        minus_A(x1,x2) = (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: plus#(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus#(plus(y,s(s(z))),plus(x,s(|0|())))
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: plus#(plus(x,s(|0|())),plus(y,s(s(z)))) -> plus#(plus(y,s(s(z))),plus(x,s(|0|())))
p2: 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,1),(0,0)) x1 + x2
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        s_A(x1) = ((0,1),(1,0)) x1 + (1,2)
        |0|_A() = (1,1)
        minus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2 + (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,1),(0,1)) x2
        plus_A(x1,x2) = x2 + (1,1)
        s_A(x1) = (2,2)
        |0|_A() = (1,1)
        minus_A(x1,x2) = x2 + (1,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
        s_A(x1) = (2,2)
        |0|_A() = (1,1)
        minus_A(x1,x2) = (1,1)
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  {p1}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  r1, r2, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2
        plus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        s_A(x1) = x1 + (1,1)
        |0|_A() = (1,1)
        minus_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = ((0,1),(0,1)) x1 + x2
        plus_A(x1,x2) = ((0,0),(1,0)) x2 + (3,1)
        s_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        |0|_A() = (1,1)
        minus_A(x1,x2) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        plus#_A(x1,x2) = (0,0)
        plus_A(x1,x2) = (1,1)
        s_A(x1) = (2,2)
        |0|_A() = (1,1)
        minus_A(x1,x2) = (1,1)
    

The next rules are strictly ordered:

  p1

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