YES

We show the termination of the TRS R:

  sum(|0|()) -> |0|()
  sum(s(x)) -> +(sum(x),s(x))
  sum1(|0|()) -> |0|()
  sum1(s(x)) -> s(+(sum1(x),+(x,x)))

-- SCC decomposition.

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

p1: sum#(s(x)) -> sum#(x)
p2: sum1#(s(x)) -> sum1#(x)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: sum1(|0|()) -> |0|()
r4: sum1(s(x)) -> s(+(sum1(x),+(x,x)))

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- Reduction pair.

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

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

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: sum1(|0|()) -> |0|()
r4: sum1(s(x)) -> s(+(sum1(x),+(x,x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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: sum1#(s(x)) -> sum1#(x)

and R consists of:

r1: sum(|0|()) -> |0|()
r2: sum(s(x)) -> +(sum(x),s(x))
r3: sum1(|0|()) -> |0|()
r4: sum1(s(x)) -> s(+(sum1(x),+(x,x)))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      sum1#_A(x1) = ((1,0),(1,0)) x1
      s_A(x1) = ((1,1),(1,1)) x1 + (1,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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