YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p6, p7}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1,x2,x3) = max{x1 + 2, x2 + 3, x3 + 9}
    a_A = 2
    b#_A(x1,x2) = max{x1 + 3, x2 + 2}
    b_A(x1,x2) = max{x1 + 7, x2 - 2}
    f_A(x1) = max{0, x1 - 8}
    c_A(x1,x2,x3) = max{x1 + 4, x2 - 1, x3 + 5}

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: c(z,x,a()) -> f(b(b(f(z),z),x))
r2: b(y,b(z,a())) -> f(b(c(f(a()),y,z),z))
r3: f(c(c(z,a(),a()),x,a())) -> 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: c#(z,x,a()) -> b#(b(f(z),z),x)
p2: b#(y,b(z,a())) -> c#(f(a()),y,z)
p3: c#(z,x,a()) -> b#(f(z),z)

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1,x2,x3) = max{x1 + 3, x2 + 2}
    a_A = 0
    b#_A(x1,x2) = max{x1 + 2, x2 + 2}
    b_A(x1,x2) = 1
    f_A(x1) = x1
    c_A(x1,x2,x3) = max{x1 + 3, x2 + 1, x3}

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1,x2,x3) = max{0, x1 - 4, x2 - 15, x3 - 11}
    a_A = 13
    b#_A(x1,x2) = max{1, x1 - 15, x2 - 16}
    b_A(x1,x2) = max{x1 + 5, x2 + 2}
    f_A(x1) = max{5, x1 - 8}
    c_A(x1,x2,x3) = max{9, x1 + 4, x2 + 7, x3}

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)