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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{69, x1 + 3}
        c_A(x1,x2,x3) = max{55, x2 + 32, x3 + 22}
        a_A = 23
        b#_A(x1,x2) = max{x1 + 1, x2 + 31}
        b_A(x1,x2) = max{38, x1 + 15, x2 + 11}
        f_A(x1) = max{33, x1 - 17}
  
    precedence:
    
      a = b = f > c = b# > f#
    
    partial status:
  
      pi(f#) = [1]
      pi(c) = []
      pi(a) = []
      pi(b#) = [1]
      pi(b) = []
      pi(f) = []

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{60, x1 - 21}
        c_A(x1,x2,x3) = max{52, x1 + 20, x2 + 50}
        a_A = 0
        b#_A(x1,x2) = max{49, x1 + 28, x2 + 26}
        b_A(x1,x2) = max{34, x1 + 32, x2 + 29}
        f_A(x1) = max{35, x1 - 19}
  
    precedence:
    
      f# = c = a = b# = b = f
    
    partial status:
  
      pi(f#) = []
      pi(c) = []
      pi(a) = []
      pi(b#) = []
      pi(b) = []
      pi(f) = []

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

  (no SCCs)