YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6}


-- Reduction pair.

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

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

and R consists of:

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

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

The next rules are strictly ordered:

  p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

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

The next rules are strictly ordered:

  p1

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