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:

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

The next rules are strictly ordered:

  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))
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)

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: +#(f(x),f(y)) -> +#(x,y)
p3: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)
p4: +#(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 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 + x2 + (2,3)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,3)
          f_A(x1) = x1 + (1,2)
  
    precedence:
    
      +# = + = f
    
    partial status:
  
      pi(+#) = [2]
      pi(+) = []
      pi(f) = [1]

The next rules are strictly ordered:

  p4

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)) -> +#(x,y)
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: +#(f(x),+(f(y),z)) -> +#(f(+(x,y)),z)
p3: +#(f(x),f(y)) -> +#(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 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 + x2 + (3,2)
          +_A(x1,x2) = ((1,0),(1,0)) x1 + x2 + (3,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (2,1)
  
    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)) -> +#(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}


-- 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)) -> +#(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 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) = x1
          +_A(x1,x2) = x1 + x2 + (1,3)
          f_A(x1) = ((1,0),(1,1)) x1 + (2,2)
  
    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: +#(f(x),f(y)) -> +#(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}


-- Reduction pair.

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

p1: +#(f(x),f(y)) -> +#(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 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) = x1
          f_A(x1) = x1 + (1,1)
  
    precedence:
    
      f > +#
    
    partial status:
  
      pi(+#) = []
      pi(f) = [1]

The next rules are strictly ordered:

  p1

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