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:

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

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: 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:

  {p1}


-- Reduction pair.

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

p1: 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 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:
          f#_A(x1,x2) = ((1,0),(0,0)) x1 + (1,1)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (2,2)
  
    precedence:
    
      f# = +
    
    partial status:
  
      pi(f#) = []
      pi(+) = [1]

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) -> +#(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          +#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (3,1)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
          a_A() = (2,2)
          b_A() = (1,1)
  
    precedence:
    
      +# > + = a = b
    
    partial status:
  
      pi(+#) = []
      pi(+) = [1]
      pi(a) = []
      pi(b) = []

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,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:

  {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: +(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          +#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
          +_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,1)
          a_A() = (2,1)
          b_A() = (1,3)
  
    precedence:
    
      +# = + = a = b
    
    partial status:
  
      pi(+#) = [1, 2]
      pi(+) = []
      pi(a) = []
      pi(b) = []

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: +#(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          +#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,4)
          a_A() = (1,2)
          +_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,2)
          b_A() = (3,1)
  
    precedence:
    
      +# = a = + = b
    
    partial status:
  
      pi(+#) = []
      pi(a) = []
      pi(+) = [1]
      pi(b) = []

The next rules are strictly ordered:

  p1

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