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:
  
      max/plus interpretations on natural numbers:
      
        a#_A(x1,x2) = max{x1 + 37, x2 + 37}
        c#_A(x1) = x1 + 22
        b_A(x1,x2) = max{15, x1}
        c_A(x1) = max{4, x1 - 5}
        |0|_A = 0
        a_A(x1,x2) = x1 + 15
  
    precedence:
    
      b = |0| = a > a# = c# = 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:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = max{1, x1 - 13}
        b_A(x1,x2) = max{29, x1, x2 - 8}
        c_A(x1) = 28
        a_A(x1,x2) = max{29, x1 + 13}
        |0|_A = 2
  
    precedence:
    
      c# = b = c = a = |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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        c#_A(x1) = max{113, x1 + 22}
        b_A(x1,x2) = max{80, x1 + 10, x2 - 13}
        c_A(x1) = max{105, x1 - 26}
        a_A(x1,x2) = max{80, x1 + 20, x2 - 51}
        |0|_A = 23
  
    precedence:
    
      c = a > c# = b = |0|
    
    partial status:
  
      pi(c#) = [1]
      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.