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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (3,3)
            f_A(x1) = ((0,0),(1,0)) x1 + (1,0)
            b_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (2,1)
            c_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x3 + (1,0)
            a_A() = (1,1)
    
      precedence:
      
        f = c > b# = b = a
      
      partial status:
    
        pi(b#) = []
        pi(f) = []
        pi(b) = [1]
        pi(c) = [3]
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = (3,3)
            f_A(x1) = (1,1)
            b_A(x1,x2) = ((0,1),(0,1)) x1 + (2,2)
            c_A(x1,x2,x3) = ((0,0),(1,1)) x3 + (1,0)
            a_A() = (1,1)
    
      precedence:
      
        b# = b = c = a > f
      
      partial status:
    
        pi(b#) = []
        pi(f) = []
        pi(b) = []
        pi(c) = []
        pi(a) = []
    

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#(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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            b#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,1)
            f_A(x1) = x1 + (1,1)
            b_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,2)
    
      precedence:
      
        b# = f = b
      
      partial status:
    
        pi(b#) = []
        pi(f) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            b#_A(x1,x2) = (0,0)
            f_A(x1) = x1 + (1,1)
            b_A(x1,x2) = x1 + (1,1)
    
      precedence:
      
        f = b > b#
      
      partial status:
    
        pi(b#) = []
        pi(f) = [1]
        pi(b) = [1]
    

The next rules are strictly ordered:

  p1

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