YES

We show the termination of the TRS R:

  f(+(x,|0|())) -> f(x)
  +(x,+(y,z)) -> +(+(x,y),z)

-- SCC decomposition.

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

p1: f#(+(x,|0|())) -> f#(x)
p2: +#(x,+(y,z)) -> +#(+(x,y),z)
p3: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3}


-- Reduction pair.

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

p1: f#(+(x,|0|())) -> f#(x)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = ((1,1),(0,0)) x1
      +_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,0)) x2 + (0,1)
      |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: +#(x,+(y,z)) -> +#(+(x,y),z)
p2: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

The set of usable rules consists of

  r2

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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