YES

We show the termination of the TRS R:

  minus(X,|0|()) -> X
  minus(s(X),s(Y)) -> p(minus(X,Y))
  p(s(X)) -> X
  div(|0|(),s(Y)) -> |0|()
  div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

-- SCC decomposition.

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

p1: minus#(s(X),s(Y)) -> p#(minus(X,Y))
p2: minus#(s(X),s(Y)) -> minus#(X,Y)
p3: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))
p4: div#(s(X),s(Y)) -> minus#(X,Y)

and R consists of:

r1: minus(X,|0|()) -> X
r2: minus(s(X),s(Y)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

The estimated dependency graph contains the following SCCs:

  {p3}
  {p2}


-- Reduction pair.

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

p1: div#(s(X),s(Y)) -> div#(minus(X,Y),s(Y))

and R consists of:

r1: minus(X,|0|()) -> X
r2: minus(s(X),s(Y)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{2, x1 + 1}
          s_A(x1) = x1 + 8
          minus_A(x1,x2) = x1 + 7
          p_A(x1) = max{15, x1 + 4}
          |0|_A = 0
    
      precedence:
      
        div# = minus > s = p = |0|
      
      partial status:
    
        pi(div#) = [1]
        pi(s) = []
        pi(minus) = []
        pi(p) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{1, x1}
          s_A(x1) = 2
          minus_A(x1,x2) = 0
          p_A(x1) = x1 + 3
          |0|_A = 0
    
      precedence:
      
        p > div# = s = minus = |0|
      
      partial status:
    
        pi(div#) = [1]
        pi(s) = []
        pi(minus) = []
        pi(p) = []
        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)) -> p(minus(X,Y))
r3: p(s(X)) -> X
r4: div(|0|(),s(Y)) -> |0|()
r5: div(s(X),s(Y)) -> s(div(minus(X,Y),s(Y)))

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.