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. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = max{x1 + 1, x2 + 1}
      *_A(x1,x2) = max{x1 + 3, x2 + 3}
      +_A(x1,x2) = max{x1, x2}
      u_A = 1
    
    2. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = 0
      *_A(x1,x2) = 0
      +_A(x1,x2) = 1
      u_A = 0
    

The next rules are strictly ordered:

  p1, p2

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)
p2: +#(+(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}


-- Reduction pair.

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

p1: +#(+(x,y),z) -> +#(y,z)
p2: +#(+(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. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = x1 + 7
      +_A(x1,x2) = max{5, x1 + 3, x2}
      *_A(x1,x2) = x1 + 2
      u_A = 4
    
    2. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = 0
      +_A(x1,x2) = 1
      *_A(x1,x2) = 0
      u_A = 0
    

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) -> +#(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. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = x1
      +_A(x1,x2) = max{x1 + 1, x2 + 1}
    
    2. max/plus interpretations on natural numbers:
    
      +#_A(x1,x2) = 0
      +_A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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