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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = ((0,1),(0,0)) x1 + (5,0)
            c_A(x1,x2,x3) = ((0,0),(1,1)) x2 + ((0,0),(0,1)) x3 + (5,0)
            a_A() = (1,1)
            b#_A(x1,x2) = ((1,1),(0,0)) x2 + (4,0)
            b_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2 + (13,2)
            f_A(x1) = ((0,1),(0,1)) x1 + (16,2)
    
      precedence:
      
        f > c > a > f# = b# = b
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = [2]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = (2,1)
            c_A(x1,x2,x3) = (4,3)
            a_A() = (3,1)
            b#_A(x1,x2) = (2,1)
            b_A(x1,x2) = (1,2)
            f_A(x1) = (1,2)
    
      precedence:
      
        a > b = f > f# = b# > c
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = []
        pi(f) = []
    

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: 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))

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#(f(z))
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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = ((1,0),(0,0)) x1 + (3,10)
            c_A(x1,x2,x3) = ((1,1),(0,0)) x2 + ((1,1),(1,1)) x3 + (0,1)
            a_A() = (1,4)
            b#_A(x1,x2) = ((0,1),(0,0)) x2 + (2,10)
            b_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (2,9)
            f_A(x1) = ((1,0),(1,0)) x1 + (7,14)
    
      precedence:
      
        f# = c = b# = f > a = b
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = [2]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = (5,3)
            c_A(x1,x2,x3) = (3,0)
            a_A() = (1,1)
            b#_A(x1,x2) = (4,2)
            b_A(x1,x2) = x2 + (3,0)
            f_A(x1) = (5,0)
    
      precedence:
      
        a > c = b = f > f# > b#
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b#) = []
        pi(b) = [2]
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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