No, this is not correct. For example, if A=1, and the rest are 0, then the second-last term from the original equation is satisfied, so Y=1, but none of your three new terms are satisfied, which would imply that Y=0.
You can only eliminate one variable at a time using this method. So your first step was correct (and the fourth step was trivial, but correct), but the second and third step were wrong.
It's a lot easier to reduce this case using a Karnaugh map. If you do, I think you'll find that it just reduces to !AC+!B!D.