How Can Boolean Algebra Simplify Complex Circuit Designs?

[Pariah]

Homework Statement

We have been given a task to develop a circuit which displays the square of a binary number on a 3 x 7 seq displays.

I have already gone through and done up the Karnaugh Maps for the task and have identified the minterms. However, I believe that these can still be simplified even more.

Homework Equations

Karnaugh Maps Output = A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'

The Attempt at a Solution

A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and C from minterms 1 and 3
a'c(b' + b) (d + d')
= a'c + a'bc'd + abcd + b'c'd' + ab'c'
Factor out B and D from minterms 2 and 3
bd(a' + a)(c' + c)
= a'c + bd + b'c'd' + ab'c'(Can I simplify this anymore?)
Can the minterm ab'c' absorb the minterm b'c'd' ?
Also by using the distributive law can I add B to the minterm a'c and then further simplify the equation?

Thank you for any assistance...

 

[Pariah]

I just had another look through and have come up with a different break down of the karnaugh maps

3. The Attempt at a Solution
A'B'CD + A'BC'D + A'BCD' + ABCD + B'C'D' + AB'C'
Factor out A' and D from minterms 1 and 2
A'D(b' + b) (c + c')
= A'D + A'BC'D + ABCD + B'C'D' + AB'C'
Factor out B and D from minterms 2 and 3
bd(a' + a)(c' + c)
= A'D + BD + B'C'D' + AB'C'

This is where I am getting stuck. Is it possible to further simplify the equation or is this the final solution?

or

Can I do the following?

Factor out D from minterms 1 and 2
D(A' + B) + B'C'D' + AB'C'
A'D + B + B'C'D' + AB'C'