Hi im new here and have never used forums like this but so far it looks very good. Im having a spot of trouble simplifying a boolean algebra expression for my homework and tried everything i can think of but nothing seems to look right.

The expression is: (p' = inverted) (++ = XOR)

(p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

Things i have tried so far:

Distributive rule:

(p'qr's') + pq's'(r'r) + pq(r'd+cd')

Cancellation ([r]*r) = 1:

(p'qr's') + pq's' + pq(r'd+cd')

XOR - (r'd+cd'):

(p'qr's') + pq's' + pq(c++d)

Then that is as far as i can get it.

I have also tried multiplying out the brackets at the beginning and using commutative law and cancellation to get to:

q'+sr'+ps'+q'rp+s'q+p+r

I would write the steps i took to get to this but its 3 pages long and more than likley i've done it all wrong.

Would appreciate any help i can get as i just can't get it into it's simplest form!

The expression is: (p' = inverted) (++ = XOR)

(p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

Things i have tried so far:

Distributive rule:

(p'qr's') + pq's'(r'r) + pq(r'd+cd')

Cancellation ([r]*r) = 1:

(p'qr's') + pq's' + pq(r'd+cd')

XOR - (r'd+cd'):

(p'qr's') + pq's' + pq(c++d)

Then that is as far as i can get it.

I have also tried multiplying out the brackets at the beginning and using commutative law and cancellation to get to:

q'+sr'+ps'+q'rp+s'q+p+r

I would write the steps i took to get to this but its 3 pages long and more than likley i've done it all wrong.

Would appreciate any help i can get as i just can't get it into it's simplest form!

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