Formula:

~(((p & q) | (r => s)) => ((p | (r => s)) & (q | (r => s))))



Clauses:

 1. p | ~r | s
 2. q | ~r | s
 3. ~p | ~q
 4. ~p | r
 5. ~p | ~s
 6. r | ~q
 7. r
 8. r | ~s
 9. ~s | ~q
10. ~s | r
11. ~s 



Refutation:

12. p | s  (1, 7)
13. q | s  (2, 7)
14. p  (11, 12)
15. q  (11, 13)
16. ~q  (3, 14)
17. []  (15, 16)



Conclusion:

-> The formula is unsatisfiable.