YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5}


-- Reduction pair.

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

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

and R consists of:

r1: f(c(a(),z,x)) -> b(a(),z)
r2: b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
r3: b(y,z) -> 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:
        
          f#_A(x1) = max{33, x1 - 32}
          c_A(x1,x2,x3) = max{x1 + 12, x2 + 47, x3 + 30}
          a_A = 0
          b#_A(x1,x2) = max{x1 + 32, x2 - 1}
          b_A(x1,x2) = max{x1 + 52, x2 + 35}
          f_A(x1) = max{6, x1 + 5}
    
      precedence:
      
        c = b = f > a > f# = b#
      
      partial status:
    
        pi(f#) = []
        pi(c) = [2, 3]
        pi(a) = []
        pi(b#) = []
        pi(b) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = 3
          c_A(x1,x2,x3) = max{7, x2 - 18}
          a_A = 28
          b#_A(x1,x2) = 3
          b_A(x1,x2) = 27
          f_A(x1) = 27
    
      precedence:
      
        a > f# = c = b# = b = f
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(c(a(),z,x)) -> b(a(),z)
r2: b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
r3: b(y,z) -> 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: f#(c(a(),z,x)) -> b#(a(),z)
p2: b#(x,b(z,y)) -> f#(b(f(f(z)),c(x,z,y)))
p3: b#(x,b(z,y)) -> f#(z)

and R consists of:

r1: f(c(a(),z,x)) -> b(a(),z)
r2: b(x,b(z,y)) -> f(b(f(f(z)),c(x,z,y)))
r3: b(y,z) -> 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:
        
          f#_A(x1) = x1 + 124
          c_A(x1,x2,x3) = max{50, x2 + 48, x3 - 1}
          a_A = 31
          b#_A(x1,x2) = max{x1 + 91, x2 + 172}
          b_A(x1,x2) = max{51, x1 + 29, x2 + 26}
          f_A(x1) = max{61, x1 - 20}
    
      precedence:
      
        c > a > b = f > f# = b#
      
      partial status:
    
        pi(f#) = [1]
        pi(c) = [2]
        pi(a) = []
        pi(b#) = [2]
        pi(b) = [1]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 43
          c_A(x1,x2,x3) = max{51, x2 + 11}
          a_A = 0
          b#_A(x1,x2) = max{93, x2 + 54}
          b_A(x1,x2) = x1 + 22
          f_A(x1) = 22
    
      precedence:
      
        f# = a = b = f > c = b#
      
      partial status:
    
        pi(f#) = []
        pi(c) = [2]
        pi(a) = []
        pi(b#) = [2]
        pi(b) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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