YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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