YES

We show the termination of the relative TRS R/S:

  R:
  g(s(x),y) -> g(f(x,y),y)

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

-- SCC decomposition.

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

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

and R consists of:

r1: g(s(x),y) -> g(f(x,y),y)
r2: f(x,y) -> x
r3: f(x,y) -> f(x,s(y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: g(s(x),y) -> g(f(x,y),y)
r2: f(x,y) -> x
r3: f(x,y) -> f(x,s(y))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        s > g > f > g#
      
      argument filter:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f) = [1]
        pi(g) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1,x2) = x1
        s_A(x1) = x1 + 1
        f_A(x1,x2) = x1 + 2
        g_A(x1,x2) = x1
    

The next rules are strictly ordered:

  p1

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