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^2
    order: standard order
    interpretations:
      dfib#_A(x1,x2) = ((0,1),(0,1)) x1
      s_A(x1) = ((1,1),(1,0)) x1 + (1,1)
      dfib_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (1,1)

The next rules are strictly ordered:

  p1, p2

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