Question

A[3] B[3] A[2] B[2] ? ? GT EQ LT Left-Most Bit Next Bit to the Right

Let's take a moment to look at the behavior of this column in the tables above. If you look at the table carefully, you can see that each of the columns can be broken into two cases:

1. The case in which the EQ value in the circuit "column" to the left is false. In this case, we know that we've already made our decision, as one of the earlier bits must have been greater than the other. We should pass each signal through to the next column, as we did in the cells marked with an arrow (→) above.The partial truth table below shows the cases of interest; fill in each of the columns below:

The case where the EQ value in the circuit "column" to the left is true.
In this case, we know that each of the bits to the left of this bit are equal, as we haven't encountered the case above.
In this case, we should compare the two input bits, as we did in the circuit above.

In this case, the relevant truth table rows should be almost identical to your truth table from the previous question:

Answer 1

By looking at the given circuit, it is very much clear that the circuit is comparing two bits whether they are greater (GT), equal (EQ) or less (LT).

Whether the bits are A[3],B[3] or A[2],B[2], it depends upon the position. Whenever we compare, we have to always go from left to right.

If A[3] and B[3] are equal, then we have to go on comparing A[2] and B[2].

CASE-1

The case in which the EQ value in the circuit "column" to the left is false

It means before A[3],B[3], the bits are not equal. But we have to fill up the given table according to the given possible combination of A[3] and B[3].

That means if A[3] and B[3] are equal, i.e. if they are 00 or 11, then EQ should be 1. Similarly, if they are 01, then LT will be 1 and if they are 10, then GT will be 1.

GT_prev	EQ_prev	LT_prev	A(3)	B(3)	GT	EQ	LT
0	0	1	0	0	0	1	0
0	0	1	0	1	0	0	1
0	0	1	1	0	1	0	0
0	0	1	1	1	0	1	0
1	0	0	0	0	0	1	0
1	0	0	0	1	0	0	1
1	0	0	1	0	1	0	0
1	0	0	1	1	0	1	0

CASE-2

The case in which the EQ value in the circuit "column" to the left is true

The table will be exactly same as above.

GT_prev	EQ_prev	LT_prev	A(3)	B(3)	GT	EQ	LT
0	1	0	0	0	0	1	0
0	1	0	0	1	0	0	1
0	1	0	1	0	1	0	0
0	1	0	1	1	0	1	0