YES
We show the termination of the TRS R:

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

and R consists of:

r1: f(s(x),y) -> f(x,s(x))
r2: f(x,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(x))
p2: f#(x,s(y)) -> f#(y,x)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = 0
        s_A(x1) = max{2, x1 - 2}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{0, x1 - 12, x2 - 17}
          s_A(x1) = max{11, x1 + 6}
          f#!_A(x1,x2) = max{1, x1 - 4, x2 - 8}
          s!_A(x1) = max{8, x1 - 1}
  
    partial status:
  
      pi(f#) = []
      pi(s) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.