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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = x1 + 4
          s_A(x1) = x1 + 5
          minus_A(x1,x2) = x1 + 4
          |0|_A = 0
    
      precedence:
      
        quot# = minus = |0| > s
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          quot#_A(x1,x2) = 2
          s_A(x1) = 1
          minus_A(x1,x2) = 3
          |0|_A = 0
    
      precedence:
      
        minus > quot# > s = |0|
      
      partial status:
    
        pi(quot#) = []
        pi(s) = []
        pi(minus) = []
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        minus# = s
      
      partial status:
    
        pi(minus#) = [2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = 0
          s_A(x1) = max{2, x1}
    
      precedence:
      
        minus# = s
      
      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:

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            plus#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + (1,1)
            minus_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,5)
            s_A(x1) = ((1,1),(1,1)) x1 + (1,4)
            |0|_A() = (2,2)
    
      precedence:
      
        plus# = minus = s = |0|
      
      partial status:
    
        pi(plus#) = []
        pi(minus) = []
        pi(s) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            plus#_A(x1,x2) = (1,1)
            minus_A(x1,x2) = (2,2)
            s_A(x1) = ((1,1),(1,1)) x1
            |0|_A() = (3,3)
    
      precedence:
      
        plus# = minus = s = |0|
      
      partial status:
    
        pi(plus#) = []
        pi(minus) = []
        pi(s) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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