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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            .#_A(0,0) = 0
            .#_A(0,1) = 0
            .#_A(1,0) = 0
            .#_A(1,1) = 0
            ._A(0,0) = 0
            ._A(0,1) = 0
            ._A(1,0) = 0
            ._A(1,1) = 0
            .#!_A(0,0) = 1
            .#!_A(0,1) = 0
            .#!_A(1,0) = 1
            .#!_A(1,1) = 0
            .!_A(0,0) = 1
            .!_A(0,1) = 1
            .!_A(1,0) = 1
            .!_A(1,1) = 1
  
    partial status:
  
      pi(.#) = [1]
      pi(.) = [1, 2]

The following rules are removed:

  p1



-- 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

  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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            .#_A(0,0) = 0
            .#_A(0,1) = 0
            .#_A(1,0) = 0
            .#_A(1,1) = 0
            ._A(0,0) = 0
            ._A(0,1) = 0
            ._A(1,0) = 0
            ._A(1,1) = 1
            .#!_A(0,0) = 1
            .#!_A(0,1) = 1
            .#!_A(1,0) = 1
            .#!_A(1,1) = 1
            .!_A(0,0) = 0
            .!_A(0,1) = 1
            .!_A(1,0) = 0
            .!_A(1,1) = 0
  
    partial status:
  
      pi(.#) = [1, 2]
      pi(.) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.