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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        p#_A(x1,x2,x3) = x1 + x2 + x3
        s_A(x1) = x1 + 1
        |0|_A() = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |0| > s > p#
      
      argument filter:
    
        pi(p#) = 3
        pi(s) = 1
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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