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 path order with precedence:
  
    precedence:
    
      dfib > s > dfib#
    
    argument filter:
  
      pi(dfib#) = [1]
      pi(s) = [1]
      pi(dfib) = [1]

The next rules are strictly ordered:

  p1, p2

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