YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2,x3) = max{0, x1 - 12, x2 - 7, x3 - 12}
            s_A(x1) = x1 + 11
            f#!_A(x1,x2,x3) = max{1, x1 - 3, x2 - 2, x3 - 7}
            s!_A(x1) = max{2, x1 - 3}
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2,x3) = max{0, x1 - 6, x2 - 4, x3 - 6}
            s_A(x1) = max{5, x1 + 2}
            f#!_A(x1,x2,x3) = max{1, x1 - 9, x2 - 8, x3 - 10}
            s!_A(x1) = max{2, x1 - 6}
        
  
    partial status:
  
      pi(f#) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.