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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      p#_A(x1,x2,x3) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((0,0,0),(1,0,0),(1,1,0)) x2 + ((0,0,0),(1,0,0),(1,0,0)) x3
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      |0|_A() = (0,1,1)

The next rules are strictly ordered:

  p1, p2

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