YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      f#_A(x1) = x1 + (2,0)
      c_A(x1,x2,x3) = ((1,1),(0,0)) x2 + ((1,1),(1,1)) x3 + (0,1)
      a_A() = (1,1)
      b#_A(x1,x2) = x1 + ((1,1),(0,0)) x2
      b_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,3)
      f_A(x1) = ((1,0),(1,0)) x1 + (2,5)

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)