YES

We show the termination of the TRS R:

  O(|0|()) -> |0|()
  +(|0|(),x) -> x
  +(x,|0|()) -> x
  +(O(x),O(y)) -> O(+(x,y))
  +(O(x),I(y)) -> I(+(x,y))
  +(I(x),O(y)) -> I(+(x,y))
  +(I(x),I(y)) -> O(+(+(x,y),I(|0|())))
  *(|0|(),x) -> |0|()
  *(x,|0|()) -> |0|()
  *(O(x),y) -> O(*(x,y))
  *(I(x),y) -> +(O(*(x,y)),y)

-- SCC decomposition.

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

p1: +#(O(x),O(y)) -> O#(+(x,y))
p2: +#(O(x),O(y)) -> +#(x,y)
p3: +#(O(x),I(y)) -> +#(x,y)
p4: +#(I(x),O(y)) -> +#(x,y)
p5: +#(I(x),I(y)) -> O#(+(+(x,y),I(|0|())))
p6: +#(I(x),I(y)) -> +#(+(x,y),I(|0|()))
p7: +#(I(x),I(y)) -> +#(x,y)
p8: *#(O(x),y) -> O#(*(x,y))
p9: *#(O(x),y) -> *#(x,y)
p10: *#(I(x),y) -> +#(O(*(x,y)),y)
p11: *#(I(x),y) -> O#(*(x,y))
p12: *#(I(x),y) -> *#(x,y)

and R consists of:

r1: O(|0|()) -> |0|()
r2: +(|0|(),x) -> x
r3: +(x,|0|()) -> x
r4: +(O(x),O(y)) -> O(+(x,y))
r5: +(O(x),I(y)) -> I(+(x,y))
r6: +(I(x),O(y)) -> I(+(x,y))
r7: +(I(x),I(y)) -> O(+(+(x,y),I(|0|())))
r8: *(|0|(),x) -> |0|()
r9: *(x,|0|()) -> |0|()
r10: *(O(x),y) -> O(*(x,y))
r11: *(I(x),y) -> +(O(*(x,y)),y)

The estimated dependency graph contains the following SCCs:

  {p9, p12}
  {p2, p3, p4, p6, p7}


-- Reduction pair.

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

p1: *#(I(x),y) -> *#(x,y)
p2: *#(O(x),y) -> *#(x,y)

and R consists of:

r1: O(|0|()) -> |0|()
r2: +(|0|(),x) -> x
r3: +(x,|0|()) -> x
r4: +(O(x),O(y)) -> O(+(x,y))
r5: +(O(x),I(y)) -> I(+(x,y))
r6: +(I(x),O(y)) -> I(+(x,y))
r7: +(I(x),I(y)) -> O(+(+(x,y),I(|0|())))
r8: *(|0|(),x) -> |0|()
r9: *(x,|0|()) -> |0|()
r10: *(O(x),y) -> O(*(x,y))
r11: *(I(x),y) -> +(O(*(x,y)),y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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: +#(O(x),O(y)) -> +#(x,y)
p2: +#(I(x),I(y)) -> +#(x,y)
p3: +#(I(x),I(y)) -> +#(+(x,y),I(|0|()))
p4: +#(O(x),I(y)) -> +#(x,y)
p5: +#(I(x),O(y)) -> +#(x,y)

and R consists of:

r1: O(|0|()) -> |0|()
r2: +(|0|(),x) -> x
r3: +(x,|0|()) -> x
r4: +(O(x),O(y)) -> O(+(x,y))
r5: +(O(x),I(y)) -> I(+(x,y))
r6: +(I(x),O(y)) -> I(+(x,y))
r7: +(I(x),I(y)) -> O(+(+(x,y),I(|0|())))
r8: *(|0|(),x) -> |0|()
r9: *(x,|0|()) -> |0|()
r10: *(O(x),y) -> O(*(x,y))
r11: *(I(x),y) -> +(O(*(x,y)),y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      +#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2
      O_A(x1) = ((1,0,0),(1,1,0),(0,0,0)) x1 + (1,2,15)
      I_A(x1) = ((1,0,0),(1,1,0),(1,0,1)) x1 + (4,5,10)
      +_A(x1,x2) = ((1,0,0),(1,0,0),(1,0,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (1,1,1)
      |0|_A() = (1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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