YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: i#(+(x,y)) -> +#(i(x),i(y))
p2: i#(+(x,y)) -> i#(x)
p3: i#(+(x,y)) -> i#(y)
p4: +#(x,+(y,z)) -> +#(+(x,y),z)
p5: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: i(|0|()) -> |0|()
r2: +(|0|(),y) -> y
r3: +(x,|0|()) -> x
r4: i(i(x)) -> x
r5: +(i(x),x) -> |0|()
r6: +(x,i(x)) -> |0|()
r7: i(+(x,y)) -> +(i(x),i(y))
r8: +(x,+(y,z)) -> +(+(x,y),z)
r9: +(+(x,i(y)),y) -> x
r10: +(+(x,y),i(y)) -> x

The estimated dependency graph contains the following SCCs:

  {p2, p3}
  {p4, p5}


-- Reduction pair.

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

p1: i#(+(x,y)) -> i#(y)
p2: i#(+(x,y)) -> i#(x)

and R consists of:

r1: i(|0|()) -> |0|()
r2: +(|0|(),y) -> y
r3: +(x,|0|()) -> x
r4: i(i(x)) -> x
r5: +(i(x),x) -> |0|()
r6: +(x,i(x)) -> |0|()
r7: i(+(x,y)) -> +(i(x),i(y))
r8: +(x,+(y,z)) -> +(+(x,y),z)
r9: +(+(x,i(y)),y) -> x
r10: +(+(x,y),i(y)) -> x

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      i#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1
      +_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + (1,1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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: i(|0|()) -> |0|()
r2: +(|0|(),y) -> y
r3: +(x,|0|()) -> x
r4: i(i(x)) -> x
r5: +(i(x),x) -> |0|()
r6: +(x,i(x)) -> |0|()
r7: i(+(x,y)) -> +(i(x),i(y))
r8: +(x,+(y,z)) -> +(+(x,y),z)
r9: +(+(x,i(y)),y) -> x
r10: +(+(x,y),i(y)) -> x

The set of usable rules consists of

  r2, r3, r5, r6, r8, r9, r10

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      +#_A(x1,x2) = ((0,0,0),(1,0,0),(1,0,0)) x1 + x2
      +_A(x1,x2) = x1 + x2 + (3,2,4)
      |0|_A() = (1,1,1)
      i_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1, p2

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