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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{120, x1 + 80, x2 + 59}
        a#_A(x1,x2) = max{89, x1 + 59, x2 + 67}
        c_A(x1) = max{23, x1 - 12}
        a_A(x1,x2) = max{40, x1 + 1, x2 + 12}
        |0|_A = 0
        b_A(x1,x2) = max{41, x1 + 25, x2 - 13}
  
    precedence:
    
      b# = a# = c = a = |0| = b
    
    partial status:
  
      pi(b#) = []
      pi(a#) = [2]
      pi(c) = []
      pi(a) = [2]
      pi(|0|) = []
      pi(b) = []

The next rules are strictly ordered:

  p3

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))
p2: a#(y,c(b(a(|0|(),x),|0|()))) -> b#(|0|(),y)
p3: a#(y,c(b(a(|0|(),x),|0|()))) -> a#(c(b(|0|(),y)),x)
p4: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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#(a(c(b(|0|(),y)),x),|0|())
p3: a#(y,c(b(a(|0|(),x),|0|()))) -> a#(c(b(|0|(),y)),x)
p4: 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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{35, x1 + 20, x2 + 2}
        a#_A(x1,x2) = max{x1 + 3, x2 + 5}
        c_A(x1) = max{20, x1 - 10}
        a_A(x1,x2) = max{20, x1, x2 + 15}
        |0|_A = 0
        b_A(x1,x2) = max{x1 + 25, x2 - 16}
  
    precedence:
    
      b# = a# = c = a = |0| = b
    
    partial status:
  
      pi(b#) = []
      pi(a#) = []
      pi(c) = []
      pi(a) = [1, 2]
      pi(|0|) = []
      pi(b) = []

The next rules are strictly ordered:

  p3

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


-- 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: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        b#_A(x1,x2) = max{152, x1 + 123, x2 + 129}
        a#_A(x1,x2) = max{x1 + 134, x2 + 97}
        c_A(x1) = max{0, x1 - 17}
        a_A(x1,x2) = max{29, x1 + 2, x2 + 25}
        |0|_A = 1
        b_A(x1,x2) = max{x1 + 48, x2 - 31}
  
    precedence:
    
      b# = a# = c = a = |0| = b
    
    partial status:
  
      pi(b#) = []
      pi(a#) = [2]
      pi(c) = []
      pi(a) = [2]
      pi(|0|) = []
      pi(b) = []

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: a#(y,c(b(a(|0|(),x),|0|()))) -> b#(|0|(),y)
p2: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)