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|())))

-- 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|())))

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|())))

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}
  {p4, p5}


-- 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|())))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          quot#_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
          s_A(x1) = x1 + (3,5)
          minus_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (2,6)
          |0|_A() = (1,0)
  
    precedence:
    
      minus = |0| > quot# = s
    
    partial status:
  
      pi(quot#) = []
      pi(s) = []
      pi(minus) = [1]
      pi(|0|) = []

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|())))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          minus#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      s > minus#
    
    partial status:
  
      pi(minus#) = []
      pi(s) = [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#(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|())))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (7,1)
          s_A(x1) = ((1,0),(1,1)) x1 + (3,13)
          minus_A(x1,x2) = x1 + (5,30)
          |0|_A() = (1,12)
  
    precedence:
    
      plus# = s = minus = |0|
    
    partial status:
  
      pi(plus#) = []
      pi(s) = [1]
      pi(minus) = [1]
      pi(|0|) = []

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#(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|())))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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|())))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          plus#_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (1,1)
          minus_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (0,1)
          s_A(x1) = ((1,0),(0,0)) x1 + (3,5)
          |0|_A() = (1,7)
  
    precedence:
    
      plus# = minus > s = |0|
    
    partial status:
  
      pi(plus#) = []
      pi(minus) = []
      pi(s) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

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