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:

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

The next rules are strictly ordered:

  p1, p2

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