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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          I_A(x1) = max{1, x1}
          O_A(x1) = max{1, x1}
    
      precedence:
      
        *# = I = O
      
      partial status:
    
        pi(*#) = [1]
        pi(I) = [1]
        pi(O) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{0, x1 - 2}
          I_A(x1) = max{3, x1 + 1}
          O_A(x1) = max{3, x1 + 1}
    
      precedence:
      
        *# = I = O
      
      partial status:
    
        pi(*#) = []
        pi(I) = [1]
        pi(O) = [1]
    

The next rules are strictly ordered:

  p1, p2

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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,2)
            O_A(x1) = ((1,0),(1,0)) x1 + (1,0)
            I_A(x1) = ((1,0),(0,0)) x1 + (9,1)
            +_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (2,4)
            |0|_A() = (5,4)
    
      precedence:
      
        +# > O = I = + > |0|
      
      partial status:
    
        pi(+#) = [2]
        pi(O) = []
        pi(I) = []
        pi(+) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = (0,0)
            O_A(x1) = (2,0)
            I_A(x1) = (1,1)
            +_A(x1,x2) = ((1,0),(1,1)) x1 + (3,2)
            |0|_A() = (3,2)
    
      precedence:
      
        O > + > +# = I = |0|
      
      partial status:
    
        pi(+#) = []
        pi(O) = []
        pi(I) = []
        pi(+) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: +#(O(x),I(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:

  {p1}


-- Reduction pair.

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

p1: +#(O(x),I(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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{0, x1 - 2, x2 - 2}
          O_A(x1) = max{3, x1 + 1}
          I_A(x1) = max{3, x1 + 1}
    
      precedence:
      
        +# = O = I
      
      partial status:
    
        pi(+#) = []
        pi(O) = [1]
        pi(I) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          O_A(x1) = max{2, x1}
          I_A(x1) = max{2, x1}
    
      precedence:
      
        +# = O = I
      
      partial status:
    
        pi(+#) = []
        pi(O) = [1]
        pi(I) = [1]
    

The next rules are strictly ordered:

  p1

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