YES

We show the termination of the TRS R:

  f(+(x,|0|())) -> f(x)
  +(x,+(y,z)) -> +(+(x,y),z)

-- SCC decomposition.

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

p1: f#(+(x,|0|())) -> f#(x)
p2: +#(x,+(y,z)) -> +#(+(x,y),z)
p3: +#(x,+(y,z)) -> +#(x,y)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,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: f#(+(x,|0|())) -> f#(x)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

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:
        
          f#_A(x1) = max{2, x1 + 1}
          +_A(x1,x2) = max{x1, x2}
          |0|_A = 0
    
      precedence:
      
        f# = + = |0|
      
      partial status:
    
        pi(f#) = [1]
        pi(+) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{1, x1 - 1}
          +_A(x1,x2) = x2
          |0|_A = 0
    
      precedence:
      
        f# = + = |0|
      
      partial status:
    
        pi(f#) = []
        pi(+) = [2]
        pi(|0|) = []
    

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)) -> +#(x,y)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

The set of usable rules consists of

  r2

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 + 3
          +_A(x1,x2) = max{x1, x2 + 3}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = []
        pi(+) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = x2 + 2
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = []
        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,y)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,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)

and R consists of:

r1: f(+(x,|0|())) -> f(x)
r2: +(x,+(y,z)) -> +(+(x,y),z)

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{0, x2 - 2}
          +_A(x1,x2) = max{x1 + 1, x2 + 3}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = []
        pi(+) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = max{x1 + 1, 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.