YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p4: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))
p5: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y)
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p8: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p9: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p10: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p11: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p12: c#(f(c(a(),y,a())),x,z) -> b#(x,a())

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p5: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y)
p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p9: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p10: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p11: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = (1,0)
          a_A() = (10,1)
          b_A(x1,x2) = ((0,1),(0,0)) x1 + (4,0)
          c_A(x1,x2,x3) = (8,3)
          c#_A(x1,x2,x3) = (1,0)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,2)
          f#_A(x1) = ((0,1),(0,0)) x1
  
    precedence:
    
      f# > a > b# = b = c = c# = f
    
    partial status:
  
      pi(b#) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(c#) = []
      pi(f) = []
      pi(f#) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p5: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p8: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p9: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p10: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p4: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p5: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))
p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x)
p7: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p9: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p10: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = (1,0)
          a_A() = (11,2)
          b_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,8)
          c_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((1,1),(0,0)) x2 + ((0,1),(1,0)) x3 + (1,3)
          c#_A(x1,x2,x3) = (1,0)
          f_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          f#_A(x1) = ((1,0),(0,0)) x1
  
    precedence:
    
      c > f > b > b# = a = c# > f#
    
    partial status:
  
      pi(b#) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(c#) = []
      pi(f) = []
      pi(f#) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))
p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a())
p4: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a())))))
p5: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y))
p6: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a()))))
p7: c#(f(c(a(),y,a())),x,z) -> b#(z,z)
p8: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a())))
p9: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p6, p7, p9}
  {p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((0,1),(0,0)) x2 + (1,0)
          a_A() = (2,1)
          b_A(x1,x2) = ((0,0),(1,0)) x1 + (0,4)
          c_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + ((1,1),(0,0)) x3 + (0,4)
          c#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + ((1,1),(0,0)) x3 + (1,0)
          f_A(x1) = ((0,1),(0,0)) x1 + (2,0)
  
    precedence:
    
      c = f > b# = a = b = c#
    
    partial status:
  
      pi(b#) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(c#) = []
      pi(f) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = x2 + (10,25)
          a_A() = (9,1)
          b_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (1,2)
          c_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2 + ((1,1),(1,1)) x3 + (31,8)
          c#_A(x1,x2,x3) = ((1,1),(1,1)) x2 + ((1,1),(0,1)) x3 + (22,28)
          f_A(x1) = (6,0)
  
    precedence:
    
      b > c = f > a = c# > b#
    
    partial status:
  
      pi(b#) = [2]
      pi(a) = []
      pi(b) = []
      pi(c) = [3]
      pi(c#) = [3]
      pi(f) = []

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((1,0),(1,0)) x2 + (2,1)
          a_A() = (0,0)
          b_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,0)
          c_A(x1,x2,x3) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + ((0,0),(1,0)) x3 + (6,0)
          c#_A(x1,x2,x3) = ((1,1),(1,1)) x2 + (10,4)
          f_A(x1) = (6,0)
  
    precedence:
    
      b# = c# > c > a = b > f
    
    partial status:
  
      pi(b#) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(c#) = [2]
      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#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x)
p2: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a()))
p3: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a()))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,3)
          a_A() = (0,2)
          b_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,0)) x2 + (1,0)
          c_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((1,0),(1,1)) x2 + ((1,1),(0,0)) x3 + (4,9)
          f_A(x1) = (0,1)
  
    precedence:
    
      c > b# = a = b > f
    
    partial status:
  
      pi(b#) = []
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(f) = []

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (1,0)
          a_A() = (1,1)
          b_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (0,2)
          c_A(x1,x2,x3) = ((0,1),(1,1)) x1 + ((0,0),(1,0)) x2 + ((0,0),(1,1)) x3 + (5,9)
          f_A(x1) = (3,5)
  
    precedence:
    
      b > b# > c = f > a
    
    partial status:
  
      pi(b#) = [2]
      pi(a) = []
      pi(b) = []
      pi(c) = []
      pi(f) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1) = ((1,0),(0,0)) x1 + (1,12)
          c_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((1,0),(1,0)) x2 + ((1,0),(0,0)) x3 + (9,23)
          a_A() = (0,26)
          b_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (4,13)
          f_A(x1) = (1,1)
  
    precedence:
    
      b > f# = c = a = f
    
    partial status:
  
      pi(f#) = []
      pi(c) = []
      pi(a) = []
      pi(b) = []
      pi(f) = []

The next rules are strictly ordered:

  p1

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