YES

We show the termination of the TRS R:

  fac(s(x)) -> *(fac(p(s(x))),s(x))
  p(s(|0|())) -> |0|()
  p(s(s(x))) -> s(p(s(x)))

-- SCC decomposition.

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

p1: fac#(s(x)) -> fac#(p(s(x)))
p2: fac#(s(x)) -> p#(s(x))
p3: p#(s(s(x))) -> p#(s(x))

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}


-- Reduction pair.

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

p1: fac#(s(x)) -> fac#(p(s(x)))

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The set of usable rules consists of

  r2, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

r1: fac(s(x)) -> *(fac(p(s(x))),s(x))
r2: p(s(|0|())) -> |0|()
r3: p(s(s(x))) -> s(p(s(x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3

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