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:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (60,65)
          a#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (40,46)
          c_A(x1) = ((0,0),(1,0)) x1 + (1,1)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(0,1)) x2 + (10,1)
          |0|_A() = (15,1)
          b_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(0,0)) x2 + (8,29)
  
    precedence:
    
      a > b > |0| > c > b# = a#
    
    partial status:
  
      pi(b#) = []
      pi(a#) = []
      pi(c) = []
      pi(a) = []
      pi(|0|) = []
      pi(b) = [1]

The next rules are strictly ordered:

  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))
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)

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

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          b#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(0,0)) x2 + (42,0)
          a#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (24,0)
          c_A(x1) = ((0,1),(0,0)) x1 + (1,1)
          a_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (2,1)
          |0|_A() = (2,11)
          b_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,0),(0,1)) x2 + (20,11)
  
    precedence:
    
      b# > a > b > |0| > a# = c
    
    partial status:
  
      pi(b#) = []
      pi(a#) = []
      pi(c) = []
      pi(a) = [1]
      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|()))) -> a#(c(b(|0|(),y)),x)
p2: a#(y,c(b(a(|0|(),x),|0|()))) -> b#(|0|(),y)
p3: b#(x,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:

  {p1, p2, p3}


-- 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)
p2: a#(y,c(b(a(|0|(),x),|0|()))) -> b#(|0|(),y)
p3: b#(x,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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1,x2) = ((1,1),(0,1)) x2 + (1,10)
          c_A(x1) = ((0,0),(1,0)) x1 + (2,1)
          b_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (3,16)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,0)
          |0|_A() = (9,0)
          b#_A(x1,x2) = ((1,1),(0,1)) x1 + (2,33)
  
    precedence:
    
      a# = c = b = a > |0| = b#
    
    partial status:
  
      pi(a#) = []
      pi(c) = []
      pi(b) = []
      pi(a) = []
      pi(|0|) = []
      pi(b#) = []

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

  {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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a#_A(x1,x2) = ((1,0),(0,0)) x2 + (84,2)
          c_A(x1) = ((0,1),(0,0)) x1 + (1,3)
          b_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (46,27)
          a_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (10,4)
          |0|_A() = (26,1)
  
    precedence:
    
      a > a# = |0| > c = b
    
    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.