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

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        .#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,0)) x2
        ._A(x1,x2) = ((1,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        .#_A(x1,x2) = ((1,1),(0,1)) x1
        ._A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

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