YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: f(b(a(),z)) -> z
r2: b(y,b(a(),z)) -> b(f(c(y,y,a())),b(f(z),a()))
r3: f(f(f(c(z,x,a())))) -> b(f(x),z)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6}


-- Reduction pair.

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

p1: b#(y,b(a(),z)) -> b#(f(c(y,y,a())),b(f(z),a()))
p2: b#(y,b(a(),z)) -> f#(z)
p3: f#(f(f(c(z,x,a())))) -> f#(x)
p4: f#(f(f(c(z,x,a())))) -> b#(f(x),z)

and R consists of:

r1: f(b(a(),z)) -> z
r2: b(y,b(a(),z)) -> b(f(c(y,y,a())),b(f(z),a()))
r3: f(f(f(c(z,x,a())))) -> b(f(x),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:
        
          b#_A(x1,x2) = max{8, x1 - 6, x2}
          b_A(x1,x2) = max{7, x1 - 7, x2 - 1}
          a_A = 0
          f_A(x1) = x1 + 6
          c_A(x1,x2,x3) = max{x1 - 6, x2 - 6, x3 + 7}
          f#_A(x1) = max{7, x1 - 1}
    
      precedence:
      
        b = f = c = f# > b# > a
      
      partial status:
    
        pi(b#) = [2]
        pi(b) = []
        pi(a) = []
        pi(f) = [1]
        pi(c) = [3]
        pi(f#) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = 58
          b_A(x1,x2) = 65
          a_A = 51
          f_A(x1) = max{57, x1 + 4}
          c_A(x1,x2,x3) = max{53, x3}
          f#_A(x1) = 56
    
      precedence:
      
        b# = a = f = c > b = f#
      
      partial status:
    
        pi(b#) = []
        pi(b) = []
        pi(a) = []
        pi(f) = [1]
        pi(c) = [3]
        pi(f#) = []
    

The next rules are strictly ordered:

  p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: b#(y,b(a(),z)) -> b#(f(c(y,y,a())),b(f(z),a()))

and R consists of:

r1: f(b(a(),z)) -> z
r2: b(y,b(a(),z)) -> b(f(c(y,y,a())),b(f(z),a()))
r3: f(f(f(c(z,x,a())))) -> b(f(x),z)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: b#(y,b(a(),z)) -> b#(f(c(y,y,a())),b(f(z),a()))

and R consists of:

r1: f(b(a(),z)) -> z
r2: b(y,b(a(),z)) -> b(f(c(y,y,a())),b(f(z),a()))
r3: f(f(f(c(z,x,a())))) -> b(f(x),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: standard order
          interpretations:
            b#_A(x1,x2) = ((0,1),(0,0)) x2
            b_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(1,0)) x2
            a_A() = (0,2)
            f_A(x1) = ((1,1),(1,0)) x1
            c_A(x1,x2,x3) = ((0,1),(1,0)) x1 + ((0,0),(1,1)) x2 + (0,1)
    
      precedence:
      
        b# = b = a = f = c
      
      partial status:
    
        pi(b#) = []
        pi(b) = []
        pi(a) = []
        pi(f) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = (0,0)
            b_A(x1,x2) = ((1,0),(1,1)) x2 + (3,3)
            a_A() = (0,0)
            f_A(x1) = (2,2)
            c_A(x1,x2,x3) = (1,1)
    
      precedence:
      
        f > b > b# = a = c
      
      partial status:
    
        pi(b#) = []
        pi(b) = [2]
        pi(a) = []
        pi(f) = []
        pi(c) = []
    

The next rules are strictly ordered:

  p1

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