YES

We show the termination of the TRS R:

  p(m,n,s(r)) -> p(m,r,n)
  p(m,s(n),|0|()) -> p(|0|(),n,m)
  p(m,|0|(),|0|()) -> m

-- SCC decomposition.

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

p1: p#(m,n,s(r)) -> p#(m,r,n)
p2: p#(m,s(n),|0|()) -> p#(|0|(),n,m)

and R consists of:

r1: p(m,n,s(r)) -> p(m,r,n)
r2: p(m,s(n),|0|()) -> p(|0|(),n,m)
r3: p(m,|0|(),|0|()) -> m

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: p#(m,n,s(r)) -> p#(m,r,n)
p2: p#(m,s(n),|0|()) -> p#(|0|(),n,m)

and R consists of:

r1: p(m,n,s(r)) -> p(m,r,n)
r2: p(m,s(n),|0|()) -> p(|0|(),n,m)
r3: p(m,|0|(),|0|()) -> m

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + x3 + (2,2)
            s_A(x1) = ((1,0),(1,1)) x1 + (3,3)
            |0|_A() = (2,1)
    
      precedence:
      
        p# = s = |0|
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (3,3)
            s_A(x1) = ((1,0),(1,0)) x1 + (2,2)
            |0|_A() = (1,1)
    
      precedence:
      
        p# > s = |0|
      
      partial status:
    
        pi(p#) = []
        pi(s) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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