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 monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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