Consider sender window size as 2 which is also the sequence number space and receiver window size is 1 always.
Now, sender sends 2 packets. Both are acknowledged by receiver but the acknowledgements are lost.
Say, timeout occurs at sender. Now, sender will send initial 0 but receiver is expecting next 0. This will create confusion.
So, we are sure that qindows size cant be equal to sequence number space.
Now, make the sender window size = 1 and see if it works.
So, sender window size <= sequnce number space - 1 for sliding window protocol