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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      a#_A(x1,x2) = ((1,1),(0,0)) x2
      c#_A(x1) = ((1,1),(0,0)) x1
      b_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (1,0)
      c_A(x1) = (1,4)
      |0|_A() = (1,1)
      a_A(x1,x2) = x1 + (2,0)

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: a#(x,y) -> c#(y)
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:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      c#_A(x1) = ((0,1),(1,1)) x1
      b_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,0)) x2 + (1,0)
      c_A(x1) = (10,1)
      a_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (1,2)
      |0|_A() = (1,1)

The next rules are strictly ordered:

  p1, p2

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