A boolean algebra manipulations (should be fun)

[thomas49th]

Homework Statement

Looking at the characteristic equation for a SR flip flop we have

Q+ = QR¬ + S

It's known (from the associated K-map of the SR/SC flippy floppy) that

(Q+)¬ = Q¬S¬ + R

K-maps are useful as they do the simplification for use, but I'm trying to use algebra to get from Q+ to (Q+)¬ (the compliment of itself)

Homework Equations

Well

(A+B)(A+C) = A + BC... maybe useful??

The Attempt at a Solution

Q+ = QR¬ + S
So (Q+)¬
= (QR¬ + S)¬ (compliment of both sides)
=> (QR¬)¬.S¬ (DeMorgan)
=> (Q¬+R)S¬
=> Q¬S¬+R¬S¬

But I cannot seem to make Q¬S¬ + R¬S¬ = Q¬S¬ + R

Any suggestions. Have I made a mistake?

Thanks
Thomas

N.B ¬ denotes compliment/inverse

 

[LCKurtz]

Homework Statement

Looking at the characteristic equation for a SR flip flop we have

Q+ = QR¬ + S

It's known (from the associated K-map of the SR/SC flippy floppy) that

(Q+)¬ = Q¬S¬ + R

K-maps are useful as they do the simplification for use, but I'm trying to use algebra to get from Q+ to (Q+)¬ (the compliment of itself)

Homework Equations

Well

(A+B)(A+C) = A + BC... maybe useful??

The Attempt at a Solution

Q+ = QR¬ + S
So (Q+)¬
= (QR¬ + S)¬ (compliment of both sides)
=> (QR¬)¬.S¬ (DeMorgan)
=> (Q¬+R)S¬
=> Q¬S¬+R¬S¬

But I cannot seem to make Q¬S¬ + R¬S¬ = Q¬S¬ + R

Any suggestions. Have I made a mistake?

Thanks
Thomas

N.B ¬ denotes compliment/inverse

It is "complement", not "compliment".

The reason you can't make them equal is that they aren't. You are forgetting about the disallowed inputs of S and R = 1 simultaneously. This gives don't care states in the transition table. So when you group the 1's with the appropriate X's and then group 0's with the X's, the X's may represent 1 in one grouping and 1 in the other grouping as well. So you aren't complementing the same function.

 

[thomas49th]

Hahah yes, I must compliment you on your wise words which have complemented by knowledge.

In short, I understand

Thanks :)