YES

We show the termination of the TRS R:

  +(x,+(y,z)) -> +(+(x,y),z)
  +(*(x,y),+(x,z)) -> *(x,+(y,z))
  +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

-- SCC decomposition.

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

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

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

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

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The set of usable rules consists of

  r1, r2, r3

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 + ((1,0),(1,0)) x2 + (4,2)
            +_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,0)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
    
      precedence:
      
        +# = + = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = [2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = (0,0)
            +_A(x1,x2) = (2,4)
            *_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1)
    
      precedence:
      
        +# = + = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: +#(x,+(y,z)) -> +#(+(x,y),z)
p2: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
p3: +#(*(x,y),+(x,z)) -> +#(y,z)
p4: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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)
p3: +#(*(x,y),+(x,z)) -> +#(y,z)
p4: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The set of usable rules consists of

  r1, r2, r3

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) = x2 + (2,4)
            +_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,2)
            *_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,3)
    
      precedence:
      
        +# = + = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = (1,1)
            +_A(x1,x2) = (2,2)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + (3,3)
    
      precedence:
      
        +# = + = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = [1]
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: +#(x,+(y,z)) -> +#(+(x,y),z)
p2: +#(*(x,y),+(x,z)) -> +#(y,z)
p3: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

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: +#(x,+(y,z)) -> +#(+(x,y),z)
p2: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
p3: +#(*(x,y),+(x,z)) -> +#(y,z)

and R consists of:

r1: +(x,+(y,z)) -> +(+(x,y),z)
r2: +(*(x,y),+(x,z)) -> *(x,+(y,z))
r3: +(*(x,y),+(*(x,z),u)) -> +(*(x,+(y,z)),u)

The set of usable rules consists of

  r1, r2, r3

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) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (4,3)
            +_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,2)
            *_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
    
      precedence:
      
        + > +# = *
      
      partial status:
    
        pi(+#) = [2]
        pi(+) = []
        pi(*) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = ((1,0),(1,0)) x2 + (4,3)
            +_A(x1,x2) = (2,2)
            *_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (3,1)
    
      precedence:
      
        +# = + = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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