YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(s(x),y) -> f#(x,s(x))
p2: f#(x,s(y)) -> f#(y,x)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(x,s(x))
p2: f#(x,s(y)) -> f#(y,x)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{x1 + 4, x2 + 2}
        s_A(x1) = x1 + 2
  
    precedence:
    
      f# > s
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{x1 + 1, x2 - 1}
        s_A(x1) = x1 + 3
  
    precedence:
    
      f# = s
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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