YES

We show the termination of the TRS R:

  .(.(x,y),z) -> .(x,.(y,z))
  a(f(x)) -> f(a(x))
  a(.(x,y)) -> .(a(x),y)
  a(b1(x)) -> b1(a(x))
  f(b(x)) -> b(f(x))
  .(b(x),y) -> b(.(x,y))
  b1(b(x)) -> b(b(x))
  a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
  a(f(|0|())) -> b1(.(f(|0|()),|0|()))
  f(.(|0|(),x)) -> b(.(|0|(),f(x)))
  f(|0|()) -> b(|0|())
  c(b(x)) -> c(a(x))
  a(b(x)) -> b(a(x))

-- SCC decomposition.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(.(x,y),z) -> .#(y,z)
p3: a#(f(x)) -> f#(a(x))
p4: a#(f(x)) -> a#(x)
p5: a#(.(x,y)) -> .#(a(x),y)
p6: a#(.(x,y)) -> a#(x)
p7: a#(b1(x)) -> b1#(a(x))
p8: a#(b1(x)) -> a#(x)
p9: f#(b(x)) -> f#(x)
p10: .#(b(x),y) -> .#(x,y)
p11: a#(f(.(|0|(),x))) -> b1#(.(f(.(|0|(),x)),.(|0|(),f(x))))
p12: a#(f(.(|0|(),x))) -> .#(f(.(|0|(),x)),.(|0|(),f(x)))
p13: a#(f(.(|0|(),x))) -> .#(|0|(),f(x))
p14: a#(f(.(|0|(),x))) -> f#(x)
p15: a#(f(|0|())) -> b1#(.(f(|0|()),|0|()))
p16: a#(f(|0|())) -> .#(f(|0|()),|0|())
p17: f#(.(|0|(),x)) -> .#(|0|(),f(x))
p18: f#(.(|0|(),x)) -> f#(x)
p19: c#(b(x)) -> c#(a(x))
p20: c#(b(x)) -> a#(x)
p21: a#(b(x)) -> a#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The estimated dependency graph contains the following SCCs:

  {p19}
  {p4, p6, p8, p21}
  {p1, p2, p10}
  {p9, p18}


-- Reduction pair.

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

p1: c#(b(x)) -> c#(a(x))

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r13

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            b_A(x1) = ((1,0),(1,0)) x1 + (2,4)
            a_A(x1) = x1 + (2,2)
            ._A(x1,x2) = x1
            f_A(x1) = x1 + (5,5)
            b1_A(x1) = ((0,1),(0,1)) x1 + (1,2)
            |0|_A() = (4,5)
    
      precedence:
      
        c# = b = a = . = f = b1 = |0|
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(a) = []
        pi(.) = []
        pi(f) = []
        pi(b1) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            b_A(x1) = ((1,1),(1,0)) x1 + (1,3)
            a_A(x1) = ((1,1),(1,0)) x1 + (0,2)
            ._A(x1,x2) = x1
            f_A(x1) = x1
            b1_A(x1) = (6,1)
            |0|_A() = (7,10)
    
      precedence:
      
        a = f = |0| > . > c# = b = b1
      
      partial status:
    
        pi(c#) = []
        pi(b) = []
        pi(a) = []
        pi(.) = []
        pi(f) = []
        pi(b1) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: a#(b(x)) -> a#(x)
p2: a#(b1(x)) -> a#(x)
p3: a#(.(x,y)) -> a#(x)
p4: a#(f(x)) -> a#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = max{2, x1 + 1}
          b_A(x1) = max{1, x1}
          b1_A(x1) = max{1, x1}
          ._A(x1,x2) = max{x1, x2}
          f_A(x1) = max{1, x1}
    
      precedence:
      
        a# = b = b1 = . = f
      
      partial status:
    
        pi(a#) = [1]
        pi(b) = [1]
        pi(b1) = [1]
        pi(.) = [1, 2]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = x1 + 3
          b_A(x1) = x1
          b1_A(x1) = x1
          ._A(x1,x2) = max{x1 + 2, x2 + 2}
          f_A(x1) = x1
    
      precedence:
      
        a# = b = b1 = . = f
      
      partial status:
    
        pi(a#) = [1]
        pi(b) = [1]
        pi(b1) = [1]
        pi(.) = [1, 2]
        pi(f) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(b(x),y) -> .#(x,y)
p3: .#(.(x,y),z) -> .#(y,z)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  r1, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          .#_A(x1,x2) = x1 + 4
          ._A(x1,x2) = max{x1 + 5, x2}
          b_A(x1) = x1
    
      precedence:
      
        .# = . > b
      
      partial status:
    
        pi(.#) = [1]
        pi(.) = [1, 2]
        pi(b) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          .#_A(x1,x2) = x1 + 3
          ._A(x1,x2) = max{x1 + 2, x2 + 5}
          b_A(x1) = x1 + 4
    
      precedence:
      
        .# = . = b
      
      partial status:
    
        pi(.#) = [1]
        pi(.) = [2]
        pi(b) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: f#(b(x)) -> f#(x)
p2: f#(.(|0|(),x)) -> f#(x)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))
r2: a(f(x)) -> f(a(x))
r3: a(.(x,y)) -> .(a(x),y)
r4: a(b1(x)) -> b1(a(x))
r5: f(b(x)) -> b(f(x))
r6: .(b(x),y) -> b(.(x,y))
r7: b1(b(x)) -> b(b(x))
r8: a(f(.(|0|(),x))) -> b1(.(f(.(|0|(),x)),.(|0|(),f(x))))
r9: a(f(|0|())) -> b1(.(f(|0|()),|0|()))
r10: f(.(|0|(),x)) -> b(.(|0|(),f(x)))
r11: f(|0|()) -> b(|0|())
r12: c(b(x)) -> c(a(x))
r13: a(b(x)) -> b(a(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{2, x1 + 1}
          b_A(x1) = max{1, x1}
          ._A(x1,x2) = max{1, x1, x2}
          |0|_A = 0
    
      precedence:
      
        f# = b = . = |0|
      
      partial status:
    
        pi(f#) = [1]
        pi(b) = [1]
        pi(.) = [1, 2]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 2
          b_A(x1) = x1 + 1
          ._A(x1,x2) = max{x1 - 1, x2 + 1}
          |0|_A = 2
    
      precedence:
      
        f# = b = . = |0|
      
      partial status:
    
        pi(f#) = []
        pi(b) = [1]
        pi(.) = [2]
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2

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