YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p4: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))
p5: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y)
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p8: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p9: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p10: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p11: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p12: c#(f(c(a(),y,a())),x,z) -> b#(x,a())

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p5: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y)
p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p9: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p10: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p11: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))

and R consists of:

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

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{x1, x2}
          a_A = 4
          b_A(x1,x2) = max{2, x1, x2}
          c_A(x1,x2,x3) = max{4, x1, x2, x3}
          c#_A(x1,x2,x3) = max{x1, x2, x3}
          f_A(x1) = max{3, x1}
          f#_A(x1) = max{1, x1}
    
      precedence:
      
        b# = a = c = c# > b = f = f#
      
      partial status:
    
        pi(b#) = []
        pi(a) = []
        pi(b) = []
        pi(c) = []
        pi(c#) = []
        pi(f) = []
        pi(f#) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = 118
          a_A = 179
          b_A(x1,x2) = 355
          c_A(x1,x2,x3) = 120
          c#_A(x1,x2,x3) = 118
          f_A(x1) = 283
          f#_A(x1) = 77
    
      precedence:
      
        a > b# = c# > b = c = f = f#
      
      partial status:
    
        pi(b#) = []
        pi(a) = []
        pi(b) = []
        pi(c) = []
        pi(c#) = []
        pi(f) = []
        pi(f#) = []
    

The next rules are strictly ordered:

  p6, p7, p10

We remove them from the problem.

-- SCC decomposition.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p5: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p6: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p7: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p8: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p7}
  {p8}


-- Reduction pair.

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

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

and R consists of:

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

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{125, x1 - 55, x2 + 41}
          a_A = 0
          b_A(x1,x2) = max{675, x1 + 145, x2 + 181}
          c_A(x1,x2,x3) = max{348, x1 + 290, x2 + 127, x3 + 308}
          c#_A(x1,x2,x3) = max{x1 + 453, x2 + 262, x3 + 126}
          f_A(x1) = max{263, x1 - 456}
    
      precedence:
      
        b = c = c# > b# = a = f
      
      partial status:
    
        pi(b#) = [2]
        pi(a) = []
        pi(b) = [1, 2]
        pi(c) = [1, 2, 3]
        pi(c#) = [2, 3]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = x2 + 4
          a_A = 6
          b_A(x1,x2) = max{x1 + 137, x2 + 101}
          c_A(x1,x2,x3) = max{x1 + 5, x2 + 7, x3 + 31}
          c#_A(x1,x2,x3) = max{x2 + 142, x3 + 104}
          f_A(x1) = 47
    
      precedence:
      
        b = c# > f > b# = a = c
      
      partial status:
    
        pi(b#) = [2]
        pi(a) = []
        pi(b) = []
        pi(c) = [3]
        pi(c#) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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

and R consists of:

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

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 + (1,8)
            c_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((1,0),(1,1)) x2 + ((0,1),(0,0)) x3 + (9,12)
            a_A() = (3,0)
            b_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,0)) x2 + (5,2)
            f_A(x1) = (4,1)
    
      precedence:
      
        f# = c = a = b = f
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(a) = []
        pi(b) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1) = (0,0)
            c_A(x1,x2,x3) = x2 + (1,2)
            a_A() = (3,3)
            b_A(x1,x2) = (1,1)
            f_A(x1) = (0,0)
    
      precedence:
      
        a > c = b > f# = f
      
      partial status:
    
        pi(f#) = []
        pi(c) = [2]
        pi(a) = []
        pi(b) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

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