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:

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

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),*(x,z)) -> +#(y,z)
p2: +#(*(x,y),+(*(x,z),u())) -> +#(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),+(*(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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          +#_A(x1,x2) = ((1,0),(1,0)) x1 + (1,10)
          *_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,5)
          +_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (3,4)
          u_A() = (0,6)
  
    precedence:
    
      + = u > +# > *
    
    partial status:
  
      pi(+#) = []
      pi(*) = [1]
      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),+(*(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}


-- 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)

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:

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

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),*(x,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:

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

The next rules are strictly ordered:

  p1

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