YES

We show the termination of the TRS R:

  +(a(),b()) -> +(b(),a())
  +(a(),+(b(),z)) -> +(b(),+(a(),z))
  +(+(x,y),z) -> +(x,+(y,z))
  f(a(),y) -> a()
  f(b(),y) -> b()
  f(+(x,y),z) -> +(f(x,z),f(y,z))

-- SCC decomposition.

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

p1: +#(a(),b()) -> +#(b(),a())
p2: +#(a(),+(b(),z)) -> +#(b(),+(a(),z))
p3: +#(a(),+(b(),z)) -> +#(a(),z)
p4: +#(+(x,y),z) -> +#(x,+(y,z))
p5: +#(+(x,y),z) -> +#(y,z)
p6: f#(+(x,y),z) -> +#(f(x,z),f(y,z))
p7: f#(+(x,y),z) -> f#(x,z)
p8: f#(+(x,y),z) -> f#(y,z)

and R consists of:

r1: +(a(),b()) -> +(b(),a())
r2: +(a(),+(b(),z)) -> +(b(),+(a(),z))
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: f(a(),y) -> a()
r5: f(b(),y) -> b()
r6: f(+(x,y),z) -> +(f(x,z),f(y,z))

The estimated dependency graph contains the following SCCs:

  {p7, p8}
  {p4, p5}
  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: +(a(),b()) -> +(b(),a())
r2: +(a(),+(b(),z)) -> +(b(),+(a(),z))
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: f(a(),y) -> a()
r5: f(b(),y) -> b()
r6: f(+(x,y),z) -> +(f(x,z),f(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,x2) = max{x1 + 1, x2 + 1}
          +_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        f# = +
      
      partial status:
    
        pi(f#) = [1]
        pi(+) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{0, x1 - 2}
          +_A(x1,x2) = max{3, x1 + 1, x2 + 1}
    
      precedence:
      
        f# = +
      
      partial status:
    
        pi(f#) = []
        pi(+) = [2]
    

The next rules are strictly ordered:

  p1, p2

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

and R consists of:

r1: +(a(),b()) -> +(b(),a())
r2: +(a(),+(b(),z)) -> +(b(),+(a(),z))
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: f(a(),y) -> a()
r5: f(b(),y) -> b()
r6: f(+(x,y),z) -> +(f(x,z),f(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 + 10, x2 + 5}
          +_A(x1,x2) = max{4, x1 + 3, x2}
          a_A = 3
          b_A = 2
    
      precedence:
      
        +# = + = a > b
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(+) = [1, 2]
        pi(a) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = x1 + 4
          +_A(x1,x2) = max{x1 + 3, x2 + 5}
          a_A = 1
          b_A = 2
    
      precedence:
      
        +# = + = a = b
      
      partial status:
    
        pi(+#) = [1]
        pi(+) = [1, 2]
        pi(a) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1, p2

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: +#(a(),+(b(),z)) -> +#(a(),z)

and R consists of:

r1: +(a(),b()) -> +(b(),a())
r2: +(a(),+(b(),z)) -> +(b(),+(a(),z))
r3: +(+(x,y),z) -> +(x,+(y,z))
r4: f(a(),y) -> a()
r5: f(b(),y) -> b()
r6: f(+(x,y),z) -> +(f(x,z),f(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{5, x1, x2 + 3}
          a_A = 1
          +_A(x1,x2) = max{x1, x2 + 2}
          b_A = 2
    
      precedence:
      
        +# = a = + = b
      
      partial status:
    
        pi(+#) = [1, 2]
        pi(a) = []
        pi(+) = [1, 2]
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          +#_A(x1,x2) = max{5, x1 + 4, x2 + 3}
          a_A = 1
          +_A(x1,x2) = max{x1 + 2, x2 + 2}
          b_A = 0
    
      precedence:
      
        +# = a = + = b
      
      partial status:
    
        pi(+#) = []
        pi(a) = []
        pi(+) = [2]
        pi(b) = []
    

The next rules are strictly ordered:

  p1

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