YES

We show the termination of the TRS R:

  +(*(x,y),*(x,z)) -> *(x,+(y,z))
  +(+(x,y),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),*(x,z)) -> +#(y,z)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),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),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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, p5}


-- Reduction pair.

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

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

and R consists of:

r1: +(*(x,y),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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) = x1 + 1
          *_A(x1,x2) = max{x1 + 5, x2 + 1}
          +_A(x1,x2) = max{x1, x2}
          u_A = 0
    
      precedence:
      
        + > +# = * = u
      
      partial status:
    
        pi(+#) = [1]
        pi(*) = []
        pi(+) = []
        pi(u) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          *_A(x1,x2) = 5
          +_A(x1,x2) = 5
          u_A = 16
    
      precedence:
      
        u > * = + > +#
      
      partial status:
    
        pi(+#) = []
        pi(*) = []
        pi(+) = []
        pi(u) = []
    

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

and R consists of:

r1: +(*(x,y),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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),*(x,z)) -> +#(y,z)
p2: +#(+(x,y),z) -> +#(x,+(y,z))
p3: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: +(*(x,y),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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 + 2
          *_A(x1,x2) = max{x1 + 10, x2 + 5}
          +_A(x1,x2) = x2
          u_A = 0
    
      precedence:
      
        +# > + > * = u
      
      partial status:
    
        pi(+#) = []
        pi(*) = [2]
        pi(+) = []
        pi(u) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          *_A(x1,x2) = max{5, x2}
          +_A(x1,x2) = 12
          u_A = 22
    
      precedence:
      
        u > + > +# = *
      
      partial status:
    
        pi(+#) = []
        pi(*) = []
        pi(+) = []
        pi(u) = []
    

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

and R consists of:

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

and R consists of:

r1: +(*(x,y),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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) = x1 + 1
          +_A(x1,x2) = max{x1 + 2, x2}
          *_A(x1,x2) = x1
          u_A = 0
    
      precedence:
      
        + > +# > * > u
      
      partial status:
    
        pi(+#) = [1]
        pi(+) = []
        pi(*) = []
        pi(u) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = 6
          *_A(x1,x2) = 6
          u_A = 11
    
      precedence:
      
        u > + > +# = *
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(*) = []
        pi(u) = []
    

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) -> +#(y,z)

and R consists of:

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

and R consists of:

r1: +(*(x,y),*(x,z)) -> *(x,+(y,z))
r2: +(+(x,y),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{x1 + 1, x2}
          +_A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(+) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1 - 1, x2 + 1}
          +_A(x1,x2) = 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.