Yes, the cards are still in increasing order from left to right after the second step.
Here is the proof :
We basically need to prove that after the 2 rearrangements have taken place, for all values of x and y,
For any row x , card in column y+1 >card in column y.
That is, for example, card in row 5 column 4 > card in row 5 column 3,etc.
I will prove this by the following example to make it easier to understand .
Example : Let us look at row 3 column 1 and 2 after the first rearrangement. Let us call the card in row 3 column 1 as m and card in row 3 column 2 as n . clearly n > m.
Now,let us do the second rearrangement. Let us say that m is in row 4 after the second rearrangement. If we can prove that no matter what, the card in row 4 column 2 now, is still greater than m, then we are done.
Case1 : n is also in row 4: Since , n> m , it is proved.
Case 2: n is above row 4: This means that the cards in row 4 column 2 ( let's call it o) will be greater than n. So o>m and hence, proved .
Case 3: n is below row 4:
Now, let us look at the cards in column 1. Let the card in column 1's row 1 be called A1, in row 2 be called A2, etc .
Also, let us look at these cards after the first rearrangement but before the second rearrangement. Let us call the corresponding cards of these cards present in column 2 as A1', A2', A3'...A10'.
What this means is that after the first rearrangement but before the second rearrangement, A1 and A1' were present in the same row and in columns 1 and 2 respectively and so on.
Now, after the second rearrangement , at most 3 of A1',A2' and A3' can be present in column 2 in the first 4 rows. We also know that n ( also called A4') is not present in the top 4 rows. This means that one of A5',A6', A7'...A10' has to be present in the top 4 rows in column 2. Now m ( also called A4) is smaller than A5, A6...A10 . This means that m is also smaller than A5',A6'...A10' . Let us assume that it is A5' that is present in these top 4 rows .
Case a) A5' is present in row 4(the same row as m ). We already know that A5' is greater than m . Hence, proved.
Case b) A5' is present in a row above row 4 . This means that the card in row 4 column 2 will be even bigger than A5'(let's call this card p) . p> A5'> m. Hence, proved .
We have, therefore, proved that for all the possible scenarios, the card in row 4 column 2 is greater than the card in row 4 column 1. We can generalise this to prove that, after the second step, for all x and y, card in row x column y+1 > card in row x column y.