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:

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

  {p2, p3}


-- 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:

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

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