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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,0)
          s_A(x1) = ((1,0),(1,1)) x1 + (3,2)
          f_A(x1,x2) = x1 + (2,1)
          g_A(x1,x2) = (1,3)
  
    precedence:
    
      g# = s = f = g
    
    partial status:
  
      pi(g#) = []
      pi(s) = []
      pi(f) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

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