YES

TRS:

  sum(0()) -> 0()
  sum(s(x)) -> +(sqr(s(x)),sum(x))
  sqr(x) -> *(x,x)
  sum(s(x)) -> +(*(s(x),s(x)),sum(x))

max/plus interpretations on N:

  sum_A(x1) = max{7, 4 + x1}
  sum#_A(x1) = max{7, 4 + x1}
  0_A = 0
  0#_A = 0
  s_A(x1) = max{1, 5 + x1}
  s#_A(x1) = max{1, 5 + x1}
  +_A(x1,x2) = max{0, 2 + x1, 1 + x2}
  +#_A(x1,x2) = max{0, 2 + x1, 1 + x2}
  sqr_A(x1) = max{6, 1 + x1}
  sqr#_A(x1) = max{6, 1 + x1}
  *_A(x1,x2) = max{6, x1, x2}
  *#_A(x1,x2) = max{6, x1, x2}

precedence:

  sqr > sum > 0 = s = + = *