YES

We show the termination of the TRS R:

  +(+(x,y),z) -> +(x,+(y,z))
  +(f(x),f(y)) -> f(+(x,y))
  +(f(x),+(f(y),z)) -> +(f(+(x,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: +#(f(x),f(y)) -> +#(x,y)
p4: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)
p5: +#(f(x),+(f(y),z)) -> +#(x,y)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

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: +#(f(x),+(f(y),z)) -> +#(x,y)
p3: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)
p4: +#(f(x),f(y)) -> +#(x,y)
p5: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

The set of usable rules consists of

  r1, r2, r3

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{x1 + 4, x2 + 4}
          +_A(x1,x2) = max{1, x1, x2}
          f_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        +# = + > f
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = 11
          f_A(x1) = max{15, x1 + 11}
    
      precedence:
      
        f > + > +#
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = []
    

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: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)
p3: +#(f(x),f(y)) -> +#(x,y)
p4: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

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) -> +#(y,z)
p3: +#(f(x),f(y)) -> +#(x,y)
p4: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

The set of usable rules consists of

  r1, r2, r3

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 + 2
          +_A(x1,x2) = x2
          f_A(x1) = max{4, x1 + 3}
    
      precedence:
      
        +# = + > f
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 0
          +_A(x1,x2) = 6
          f_A(x1) = x1 + 8
    
      precedence:
      
        f > + > +#
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = [1]
    

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

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(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: +#(+(x,y),z) -> +#(x,+(y,z))
p2: +#(+(x,y),z) -> +#(y,z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x1 + 6
          +_A(x1,x2) = max{3, x1 + 2, x2}
          f_A(x1) = 0
    
      precedence:
      
        +# > + = f
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = 2
          +_A(x1,x2) = 12
          f_A(x1) = 6
    
      precedence:
      
        + > +# = f
      
      partial status:
    
        pi(+#) = []
        pi(+) = []
        pi(f) = []
    

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) -> +#(y,z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(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) -> +#(y,z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(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{x1 + 1, x2}
          +_A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        +# = +
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(+) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{x1 - 1, x2 + 1}
          +_A(x1,x2) = 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.

-- Reduction pair.

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

p1: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)

and R consists of:

r1: +(+(x,y),z) -> +(x,+(y,z))
r2: +(f(x),f(y)) -> f(+(x,y))
r3: +(f(x),+(f(y),z)) -> +(f(+(x,y)),z)

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = x2 + (4,2)
            f_A(x1) = (3,1)
            +_A(x1,x2) = x1 + x2 + (2,2)
    
      precedence:
      
        +# = f = +
      
      partial status:
    
        pi(+#) = []
        pi(f) = []
        pi(+) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            +#_A(x1,x2) = ((1,0),(0,0)) x2 + (1,2)
            f_A(x1) = (2,1)
            +_A(x1,x2) = x1 + x2 + (0,3)
    
      precedence:
      
        +# = f = +
      
      partial status:
    
        pi(+#) = []
        pi(f) = []
        pi(+) = []
    

The next rules are strictly ordered:

  p1

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