YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p3}
  {p1, p5, p6}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, 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{4, x1 + 2}
          c_A(x1,x2,x3) = max{x1 + 5, x2 + 5, x3}
          b_A(x1,x2) = max{x1 + 5, x2}
          a_A = 4
          f_A(x1) = max{15, x1}
    
      precedence:
      
        c = b = a > f# > f
      
      partial status:
    
        pi(f#) = [1]
        pi(c) = [3]
        pi(b) = [2]
        pi(a) = []
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{11, x1 - 3}
          c_A(x1,x2,x3) = x3 + 36
          b_A(x1,x2) = x2 + 8
          a_A = 9
          f_A(x1) = x1 + 10
    
      precedence:
      
        b > f# = c = a = f
      
      partial status:
    
        pi(f#) = []
        pi(c) = [3]
        pi(b) = []
        pi(a) = []
        pi(f) = [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: c#(c(z,y,a()),a(),a()) -> b#(z,y)
p2: b#(z,b(c(a(),y,a()),f(f(x)))) -> c#(y,a(),z)
p3: b#(z,b(c(a(),y,a()),f(f(x)))) -> c#(c(y,a(),z),z,x)

and R consists of:

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

The set of usable rules consists of

  r1, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1,x2,x3) = max{x1 + 75, x2 + 1, x3 + 106}
          c_A(x1,x2,x3) = max{x1 + 32, x2 + 45, x3 + 28}
          a_A = 17
          b#_A(x1,x2) = max{x1 + 107, x2 + 32}
          b_A(x1,x2) = max{x1 + 61, x2 + 77}
          f_A(x1) = max{18, x1 + 9}
    
      precedence:
      
        c# = c = a = b# = b = f
      
      partial status:
    
        pi(c#) = [1, 2]
        pi(c) = [1, 2, 3]
        pi(a) = []
        pi(b#) = [1, 2]
        pi(b) = [1, 2]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1,x2,x3) = max{0, x1 - 1, x2 - 1}
          c_A(x1,x2,x3) = max{x1 + 9, x2 + 8, x3 + 6}
          a_A = 4
          b#_A(x1,x2) = max{x1 + 5, x2 + 2}
          b_A(x1,x2) = max{x1 + 22, x2 + 18}
          f_A(x1) = x1 + 5
    
      precedence:
      
        c# = c = a = b# = b = f
      
      partial status:
    
        pi(c#) = []
        pi(c) = [2, 3]
        pi(a) = []
        pi(b#) = [1, 2]
        pi(b) = [2]
        pi(f) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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