YES

We show the termination of the TRS R:

  a(x,y) -> b(x,b(|0|(),c(y)))
  c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
  b(y,|0|()) -> y

-- SCC decomposition.

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

p1: a#(x,y) -> b#(x,b(|0|(),c(y)))
p2: a#(x,y) -> b#(|0|(),c(y))
p3: a#(x,y) -> c#(y)
p4: c#(b(y,c(x))) -> c#(c(b(a(|0|(),|0|()),y)))
p5: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))
p6: c#(b(y,c(x))) -> b#(a(|0|(),|0|()),y)
p7: c#(b(y,c(x))) -> a#(|0|(),|0|())

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The estimated dependency graph contains the following SCCs:

  {p3, p4, p5, p7}


-- Reduction pair.

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

p1: a#(x,y) -> c#(y)
p2: c#(b(y,c(x))) -> a#(|0|(),|0|())
p3: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))
p4: c#(b(y,c(x))) -> c#(c(b(a(|0|(),|0|()),y)))

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{x1 + 7, x2 + 18}
          c#_A(x1) = x1 + 13
          b_A(x1,x2) = max{36, x1}
          c_A(x1) = 22
          |0|_A = 0
          a_A(x1,x2) = x1 + 36
    
      precedence:
      
        a > b = c > a# > c# = |0|
      
      partial status:
    
        pi(a#) = [1, 2]
        pi(c#) = []
        pi(b) = [1]
        pi(c) = []
        pi(|0|) = []
        pi(a) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1,x2) = max{x1 + 150, x2 + 33}
          c#_A(x1) = 0
          b_A(x1,x2) = max{47, x1 + 6}
          c_A(x1) = 71
          |0|_A = 10
          a_A(x1,x2) = max{60, x1 + 9}
    
      precedence:
      
        c > a > b > |0| > a# = c#
      
      partial status:
    
        pi(a#) = [2]
        pi(c#) = []
        pi(b) = [1]
        pi(c) = []
        pi(|0|) = []
        pi(a) = []
    

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: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))
p2: c#(b(y,c(x))) -> c#(c(b(a(|0|(),|0|()),y)))

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))
p2: c#(b(y,c(x))) -> c#(c(b(a(|0|(),|0|()),y)))

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 24
          b_A(x1,x2) = max{8, x1, x2 - 2}
          c_A(x1) = max{4, x1 - 11}
          a_A(x1,x2) = max{x1 + 8, x2 - 12}
          |0|_A = 0
    
      precedence:
      
        a > b = c > c# = |0|
      
      partial status:
    
        pi(c#) = []
        pi(b) = [1]
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = 4
          b_A(x1,x2) = 47
          c_A(x1) = 54
          a_A(x1,x2) = 47
          |0|_A = 20
    
      precedence:
      
        b = c = a > c# = |0|
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(c) = []
        pi(a) = []
        pi(|0|) = []
    

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: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: c#(b(y,c(x))) -> c#(b(a(|0|(),|0|()),y))

and R consists of:

r1: a(x,y) -> b(x,b(|0|(),c(y)))
r2: c(b(y,c(x))) -> c(c(b(a(|0|(),|0|()),y)))
r3: b(y,|0|()) -> y

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{35, x1 + 22}
          b_A(x1,x2) = max{34, x1 + 16, x2 - 75}
          c_A(x1) = 168
          a_A(x1,x2) = max{x1 + 38, x2 - 62}
          |0|_A = 37
    
      precedence:
      
        c# > a > b = c = |0|
      
      partial status:
    
        pi(c#) = [1]
        pi(b) = []
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = 12
          b_A(x1,x2) = 14
          c_A(x1) = 16
          a_A(x1,x2) = x1 + 24
          |0|_A = 7
    
      precedence:
      
        a = |0| > c# = b = c
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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