YES
We show the termination of the TRS R:

  g(x,y) -> x
  g(x,y) -> y
  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: g(x,y) -> x
r2: g(x,y) -> y
r3: 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: g(x,y) -> x
r2: g(x,y) -> y
r3: f(s(x),y,y) -> f(y,x,s(x))

The set of usable rules consists of

  r1, r2

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!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2,x3) = 0
              s_A(x1) = x1 + 2
              g_A(x1,x2) = x1 + x2
              f#!_A(x1,x2,x3) = x1 + x2
              s!_A(x1) = 1
              g!_A(x1,x2) = x1 + x2
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2,x3) = 0
              s_A(x1) = x1 + 1
              g_A(x1,x2) = x1 + x2
              f#!_A(x1,x2,x3) = x2 + 1
              s!_A(x1) = 0
              g!_A(x1,x2) = x1 + x2
        
  
    partial status:
  
      pi(f#) = [3]
      pi(s) = []
      pi(g) = [1, 2]

The following rules are removed:

  p1

Then no dependency pair remains.