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:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        .#_A(x1,x2) = max{x1 + 8, x2 + 6}
        ._A(x1,x2) = max{4, x1 + 1, x2}
  
    base order 2:
  
      precedence:
      
        .# = .
      
  
    partial status:
  
      pi(.#) = [1]
      pi(.) = [1, 2]

The following rules are removed:

  p2



-- SCC decomposition.

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

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

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        .#_A(x1,x2) = max{x1 + 9, x2 - 7}
        ._A(x1,x2) = max{9, x1 + 5, x2 - 10}
  
    base order 2:
  
      precedence:
      
        .# = .
      
  
    partial status:
  
      pi(.#) = [1]
      pi(.) = []

The following rules are removed:

  p1

Then no dependency pair remains.