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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        dfib#_A(x1,x2) = x1 + 1
        s_A(x1) = max{4, x1 + 3}
        dfib_A(x1,x2) = max{2, x1 - 7}
  
    precedence:
    
      dfib# = s = dfib
    
    partial status:
  
      pi(dfib#) = [1]
      pi(s) = [1]
      pi(dfib) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        dfib#_A(x1,x2) = max{0, x1 - 3}
        s_A(x1) = max{4, x1 + 1}
  
    precedence:
    
      dfib# = s
    
    partial status:
  
      pi(dfib#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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