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

  r1

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:
  
      model-based reduction triple (=A, >=A!, >A!):
      
        finite algebra:
        
          carrier: {0,1}
          interpretations:
            f#_A(0,0,0) = 0
            f#_A(0,0,1) = 0
            f#_A(0,1,0) = 0
            f#_A(0,1,1) = 0
            f#_A(1,0,0) = 0
            f#_A(1,0,1) = 0
            f#_A(1,1,0) = 0
            f#_A(1,1,1) = 0
            s_A(0) = 1
            s_A(1) = 0
            f_A(0,0,0) = 0
            f_A(0,0,1) = 0
            f_A(0,1,0) = 0
            f_A(0,1,1) = 0
            f_A(1,0,0) = 0
            f_A(1,0,1) = 0
            f_A(1,1,0) = 0
            f_A(1,1,1) = 0
            f#!_A(0,0,0) = 1
            f#!_A(0,0,1) = 0
            f#!_A(0,1,0) = 0
            f#!_A(0,1,1) = 1
            f#!_A(1,0,0) = 1
            f#!_A(1,0,1) = 0
            f#!_A(1,1,0) = 0
            f#!_A(1,1,1) = 1
            s!_A(0) = 1
            s!_A(1) = 1
            f!_A(0,0,0) = 1
            f!_A(0,0,1) = 0
            f!_A(0,1,0) = 0
            f!_A(0,1,1) = 1
            f!_A(1,0,0) = 1
            f!_A(1,0,1) = 0
            f!_A(1,1,0) = 0
            f!_A(1,1,1) = 1
  
    partial status:
  
      pi(f#) = []
      pi(s) = [1]
      pi(f) = []

The following rules are removed:

  p1

Then no dependency pair remains.