YES

We show the termination of the TRS R:

  .(.(x,y),z) -> .(x,.(y,z))

-- 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)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))

The set of usable rules consists of

  r1

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}
    
      precedence:
      
        .# = .
      
      partial status:
    
        pi(.#) = []
        pi(.) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          .#_A(x1,x2) = 0
          ._A(x1,x2) = 2
    
      precedence:
      
        . > .#
      
      partial status:
    
        pi(.#) = []
        pi(.) = []
    

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: .#(.(x,y),z) -> .#(y,z)

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: .(.(x,y),z) -> .(x,.(y,z))

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:
        
          .#_A(x1,x2) = max{x1 + 1, x2}
          ._A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        .# = .
      
      partial status:
    
        pi(.#) = [1, 2]
        pi(.) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          .#_A(x1,x2) = max{x1 - 1, x2 + 1}
          ._A(x1,x2) = x2
    
      precedence:
      
        .# = .
      
      partial status:
    
        pi(.#) = []
        pi(.) = [2]
    

The next rules are strictly ordered:

  p1

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