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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = (2,1)
        a_A() = (1,1)
        b_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,1)
        c_A(x1,x2,x3) = ((1,1),(0,0)) x1 + ((0,0),(1,1)) x2 + ((0,1),(1,0)) x3 + (1,1)
        c#_A(x1,x2,x3) = (2,1)
        f_A(x1) = x1
        f#_A(x1) = ((0,1),(0,1)) x1
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = (0,0)
        a_A() = (1,1)
        b_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (1,0)
        c_A(x1,x2,x3) = (1,1)
        c#_A(x1,x2,x3) = (0,0)
        f_A(x1) = (2,0)
        f#_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = (1,0)
        a_A() = (1,1)
        b_A(x1,x2) = (0,1)
        c_A(x1,x2,x3) = (1,1)
        c#_A(x1,x2,x3) = (1,0)
        f_A(x1) = (2,2)
        f#_A(x1) = (0,1)
    

The next rules are strictly ordered:

  p5, p6, p7, p10, p11

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) -> b#(z,z)
p6: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- 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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = x2
        a_A() = (0,1)
        b_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,1)
        c_A(x1,x2,x3) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2 + ((1,1),(1,0)) x3 + (1,2)
        c#_A(x1,x2,x3) = ((1,1),(0,0)) x2 + x3 + (2,0)
        f_A(x1) = ((0,1),(0,0)) x1 + (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = x2
        a_A() = (1,1)
        b_A(x1,x2) = (1,1)
        c_A(x1,x2,x3) = (0,1)
        c#_A(x1,x2,x3) = x3 + (2,1)
        f_A(x1) = (2,2)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        b#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,1)
        a_A() = (1,1)
        b_A(x1,x2) = (1,1)
        c_A(x1,x2,x3) = (1,1)
        c#_A(x1,x2,x3) = (0,0)
        f_A(x1) = (2,0)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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