YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

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:
        
          b#_A(x1,x2) = max{535, x1 + 138, x2 + 133}
          a#_A(x1,x2) = max{x1 + 134, x2 + 16}
          c_A(x1) = max{401, x1 - 78}
          a_A(x1,x2) = max{379, x1 + 5, x2 + 115}
          |0|_A = 132
          b_A(x1,x2) = max{416, x1 + 228, x2 - 151}
    
      precedence:
      
        |0| > b# = a = b > a# = c
      
      partial status:
    
        pi(b#) = [1, 2]
        pi(a#) = [2]
        pi(c) = []
        pi(a) = [1, 2]
        pi(|0|) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1,x2) = max{x1 + 168, x2}
          a#_A(x1,x2) = x2 + 125
          c_A(x1) = 27
          a_A(x1,x2) = x1 + 42
          |0|_A = 0
          b_A(x1,x2) = 42
    
      precedence:
      
        a = b > c > b# = a# > |0|
      
      partial status:
    
        pi(b#) = [2]
        pi(a#) = []
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p2, p3, p4, p5

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)