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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        +_A(x1,x2) = x1 + x2 + 1
        f_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        +_A(x1,x2) = x1 + x2 + 1
        f_A(x1) = 1
    

The next rules are strictly ordered:

  p2, p3, p4, p5

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,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) -> +#(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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1
        +_A(x1,x2) = x1
        f_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1
        +_A(x1,x2) = x1 + 1
        f_A(x1) = 1
    

The next rules are strictly ordered:

  p1

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