YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{x1 - 3, x2 + 5}
        b_A(x1,x2) = max{22, x1 + 14, x2 + 18}
        c_A(x1,x2,x3) = max{x1 + 38, x2 + 34, x3 + 37}
        a_A = 0
        c#_A(x1,x2,x3) = max{x1 + 15, x2 + 10, x3 + 11}
        f_A(x1) = max{4, x1 - 6}
  
    precedence:
    
      b# = b = c = a = c# = f
    
    partial status:
  
      pi(b#) = []
      pi(b) = [2]
      pi(c) = [3]
      pi(a) = []
      pi(c#) = []
      pi(f) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{10, x1 - 1, x2 - 1}
        b_A(x1,x2) = max{x1 + 26, x2 + 30}
        c_A(x1,x2,x3) = max{x1 + 55, x2 + 59, x3 + 53}
        a_A = 0
        c#_A(x1,x2,x3) = max{x1 + 28, x2 + 24, x3 + 28}
        f_A(x1) = max{84, x1 - 7}
  
    precedence:
    
      b# = b = c = a = c# = f
    
    partial status:
  
      pi(b#) = []
      pi(b) = [2]
      pi(c) = [3]
      pi(a) = []
      pi(c#) = []
      pi(f) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)