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, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            .#_A(x1,x2) = 0
            ._A(x1,x2) = max{1, x1}
            .#!_A(x1,x2) = 0
            .!_A(x1,x2) = max{2, x2 + 1}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              .#_A(x1,x2) = x1 + x2
              ._A(x1,x2) = x1 + x2 + 1
              .#!_A(x1,x2) = x1
              .!_A(x1,x2) = x1 + 2
        
  
    partial status:
  
      pi(.#) = []
      pi(.) = [1]

The following rules are removed:

  p1, p2

Then no dependency pair remains.