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:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x2 + 8
          +_A(x1,x2) = max{x1, x2 + 9}
          *_A(x1,x2) = max{x1 + 18, x2 + 6}
    
      precedence:
      
        +# = + > *
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
        pi(*) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = x2 + 4
          *_A(x1,x2) = max{9, x1 + 6}
    
      precedence:
      
        * > +# = +
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
        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:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x2 + 8
          +_A(x1,x2) = max{x1, x2 + 4}
          *_A(x1,x2) = max{x1 + 4, x2 + 3}
    
      precedence:
      
        + > +# > *
      
      partial status:
    
        pi(+#) = [2]
        pi(+) = []
        pi(*) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 1
          +_A(x1,x2) = 1
          *_A(x1,x2) = max{1, x2}
    
      precedence:
      
        +# > + > *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = []
    

The next rules are strictly ordered:

  p3

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,z)) -> +#(x,y)
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,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:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x2 + 26
          +_A(x1,x2) = max{5, x1, x2 + 4}
          *_A(x1,x2) = max{18, x1 + 8, x2 + 2}
    
      precedence:
      
        + > +# = *
      
      partial status:
    
        pi(+#) = [2]
        pi(+) = []
        pi(*) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = 3
          *_A(x1,x2) = max{7, x1 + 4, x2}
    
      precedence:
      
        + = * > +#
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = [2]
    

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,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}


-- 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: +(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:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x2 + 4
          +_A(x1,x2) = max{3, x1, x2 + 2}
          *_A(x1,x2) = max{x1 + 3, x2 + 2}
    
      precedence:
      
        +# = + > *
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
        pi(*) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = x2 + 4
          *_A(x1,x2) = max{x1 + 13, x2 + 10}
    
      precedence:
      
        * > + > +#
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
        pi(*) = [2]
    

The next rules are strictly ordered:

  p1

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)

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}


-- Reduction pair.

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

p1: +#(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

  (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, x2 - 2}
          +_A(x1,x2) = max{x1 + 1, x2 + 3}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = []
        pi(+) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = max{x1 + 1, x2}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
    

The next rules are strictly ordered:

  p1

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