YES

We show the termination of the TRS R:

  +(x,|0|()) -> x
  +(x,i(x)) -> |0|()
  +(+(x,y),z) -> +(x,+(y,z))
  *(x,+(y,z)) -> +(*(x,y),*(x,z))
  *(+(x,y),z) -> +(*(x,z),*(y,z))

-- 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)
p3: *#(x,+(y,z)) -> +#(*(x,y),*(x,z))
p4: *#(x,+(y,z)) -> *#(x,y)
p5: *#(x,+(y,z)) -> *#(x,z)
p6: *#(+(x,y),z) -> +#(*(x,z),*(y,z))
p7: *#(+(x,y),z) -> *#(x,z)
p8: *#(+(x,y),z) -> *#(y,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

The estimated dependency graph contains the following SCCs:

  {p4, p5, p7, p8}
  {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,z)
p3: *#(x,+(y,z)) -> *#(x,z)
p4: *#(x,+(y,z)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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),(1,1)) x2 + (2,2)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      *# = +
    
    partial status:
  
      pi(*#) = [2]
      pi(+) = [1, 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,z)
p2: *#(x,+(y,z)) -> *#(x,z)
p3: *#(x,+(y,z)) -> *#(x,y)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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) = ((0,0),(1,0)) x1 + (1,1)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + (2,2)
  
    precedence:
    
      *# = +
    
    partial status:
  
      pi(*#) = []
      pi(+) = [1]

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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) = x2 + (1,1)
          +_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,2)
  
    precedence:
    
      *# = +
    
    partial status:
  
      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,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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,z)

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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),(1,0)) x2 + (1,0)
          +_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,0)
  
    precedence:
    
      + > *#
    
    partial status:
  
      pi(*#) = []
      pi(+) = []

The next rules are strictly ordered:

  p1

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

-- 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,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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),(0,0)) x1
          +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
          |0|_A() = (1,1)
          i_A(x1) = ((1,0),(1,1)) x1 + (2,2)
  
    precedence:
    
      +# = + = |0| = i
    
    partial status:
  
      pi(+#) = []
      pi(+) = [1]
      pi(|0|) = []
      pi(i) = [1]

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))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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,z))

and R consists of:

r1: +(x,|0|()) -> x
r2: +(x,i(x)) -> |0|()
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: *(x,+(y,z)) -> +(*(x,y),*(x,z))
r5: *(+(x,y),z) -> +(*(x,z),*(y,z))

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 + ((0,0),(1,0)) x2 + (3,1)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
          |0|_A() = (2,2)
          i_A(x1) = ((1,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      +# = + = |0| = i
    
    partial status:
  
      pi(+#) = []
      pi(+) = [1]
      pi(|0|) = []
      pi(i) = []

The next rules are strictly ordered:

  p1

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