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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2,x3) = max{x1 + 4, x2 + 6, x3 - 1}
        s_A(x1) = x1 + 7
        |0|_A = 3
  
    precedence:
    
      p# = s = |0|
    
    partial status:
  
      pi(p#) = []
      pi(s) = [1]
      pi(|0|) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

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}


-- Reduction pair.

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

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

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        p#_A(x1,x2,x3) = max{0, x2 - 2, x3 - 3}
        s_A(x1) = max{4, x1 + 2}
  
    precedence:
    
      p# = s
    
    partial status:
  
      pi(p#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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