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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      .#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x2
      ._A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,1,1)) x2 + (1,0,2,3)

The next rules are strictly ordered:

  p1, p2

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