YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: f(s(x),y) -> f(x,s(s(x)))
r2: f(x,s(s(y))) -> f(y,x)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: f(s(x),y) -> f(x,s(s(x)))
r2: f(x,s(s(y))) -> f(y,x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = max{x1 + 2, x2 + 1}
    s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(s(x),y) -> f#(x,s(s(x)))

and R consists of:

r1: f(s(x),y) -> f(x,s(s(x)))
r2: f(x,s(s(y))) -> f(y,x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(x,s(s(x)))

and R consists of:

r1: f(s(x),y) -> f(x,s(s(x)))
r2: f(x,s(s(y))) -> f(y,x)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = max{4, x1, x2 - 4}
    s_A(x1) = max{5, x1 + 2}

The next rules are strictly ordered:

  p1

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