YES

We show the termination of the TRS R:

  dfib(s(s(x)),y) -> dfib(s(x),dfib(x,y))

-- SCC decomposition.

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

p1: dfib#(s(s(x)),y) -> dfib#(s(x),dfib(x,y))
p2: dfib#(s(s(x)),y) -> dfib#(x,y)

and R consists of:

r1: dfib(s(s(x)),y) -> dfib(s(x),dfib(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: dfib#(s(s(x)),y) -> dfib#(s(x),dfib(x,y))
p2: dfib#(s(s(x)),y) -> dfib#(x,y)

and R consists of:

r1: dfib(s(s(x)),y) -> dfib(s(x),dfib(x,y))

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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