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:

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

The next rules are strictly ordered:

  p4

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)) -> +#(y,z)
p3: +#(*(x,y),+(*(x,z),u)) -> +#(*(x,+(y,z)),u)
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),u)) -> +#(*(x,+(y,z)),u)
p4: +#(*(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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p4

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = x2 + 10
        +_A(x1,x2) = max{x1, x2 + 10}
        *_A(x1,x2) = max{x1 + 8, x2 - 2}
  
    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,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:

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

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:

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

The next rules are strictly ordered:

  p1

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