YES

We show the termination of the TRS R:

  -(|0|(),y) -> |0|()
  -(x,|0|()) -> x
  -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
  p(|0|()) -> |0|()
  p(s(x)) -> x

-- SCC decomposition.

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

p1: -#(x,s(y)) -> -#(x,p(s(y)))
p2: -#(x,s(y)) -> p#(s(y))

and R consists of:

r1: -(|0|(),y) -> |0|()
r2: -(x,|0|()) -> x
r3: -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
r4: p(|0|()) -> |0|()
r5: p(s(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: -#(x,s(y)) -> -#(x,p(s(y)))

and R consists of:

r1: -(|0|(),y) -> |0|()
r2: -(x,|0|()) -> x
r3: -(x,s(y)) -> if(greater(x,s(y)),s(-(x,p(s(y)))),|0|())
r4: p(|0|()) -> |0|()
r5: p(s(x)) -> x

The set of usable rules consists of

  r5

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      -#_A(x1,x2) = ((1,1),(0,1)) x2
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)
      p_A(x1) = ((0,1),(0,1)) x1

The next rules are strictly ordered:

  p1

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