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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,1),(0,0)) x2 + (77,0)
            a#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (9,0)
            c_A(x1) = ((0,1),(0,0)) x1 + (1,10)
            a_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (14,0)
            |0|_A() = (42,1)
            b_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (70,32)
    
      precedence:
      
        a# = a = |0| > b# = c = b
      
      partial status:
    
        pi(b#) = []
        pi(a#) = []
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = (0,0)
            a#_A(x1,x2) = (0,0)
            c_A(x1) = (2,2)
            a_A(x1,x2) = x1 + (1,1)
            |0|_A() = (2,2)
            b_A(x1,x2) = (3,2)
    
      precedence:
      
        |0| > a > b > b# = a# = c
      
      partial status:
    
        pi(b#) = []
        pi(a#) = []
        pi(c) = []
        pi(a) = [1]
        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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (6,0)
            a#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (5,0)
            c_A(x1) = ((0,1),(0,0)) x1
            a_A(x1,x2) = x1 + ((0,0),(1,0)) x2
            |0|_A() = (0,2)
            b_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (0,1)
    
      precedence:
      
        b# = a = b > c > |0| > a#
      
      partial status:
    
        pi(b#) = []
        pi(a#) = []
        pi(c) = []
        pi(a) = [1]
        pi(|0|) = []
        pi(b) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            b#_A(x1,x2) = (0,0)
            a#_A(x1,x2) = (1,0)
            c_A(x1) = (2,2)
            a_A(x1,x2) = ((1,0),(1,0)) x1 + (4,4)
            |0|_A() = (0,1)
            b_A(x1,x2) = (3,3)
    
      precedence:
      
        c = b > a# = a > |0| > b#
      
      partial status:
    
        pi(b#) = []
        pi(a#) = []
        pi(c) = []
        pi(a) = []
        pi(|0|) = []
        pi(b) = []
    

The next rules are strictly ordered:

  p1, p2

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


-- Reduction pair.

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

p1: a#(y,c(b(a(|0|(),x),|0|()))) -> a#(c(b(|0|(),y)),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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a#_A(x1,x2) = ((0,1),(0,1)) x2 + (7,13)
            c_A(x1) = ((0,0),(1,0)) x1 + (0,1)
            b_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (0,11)
            a_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (1,3)
            |0|_A() = (0,5)
    
      precedence:
      
        a = |0| > a# = c = b
      
      partial status:
    
        pi(a#) = []
        pi(c) = []
        pi(b) = []
        pi(a) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a#_A(x1,x2) = (0,0)
            c_A(x1) = (2,0)
            b_A(x1,x2) = (3,0)
            a_A(x1,x2) = (4,1)
            |0|_A() = (1,0)
    
      precedence:
      
        b > a# = c > a = |0|
      
      partial status:
    
        pi(a#) = []
        pi(c) = []
        pi(b) = []
        pi(a) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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