Quick boolean algebra problem..help needed

techguy

HI,
I hope this is the right forum..
Well I am just doing a bit of revising for boolean algebra and de morgans rule and came across this in an equation.. I cant seem to work out how part 1 = part 2 or if it is even correct? COuld someone please shed some light on it? See attachment.

Thanks,
Hakko

Webmonkey

I actually don't think those boolean expressions are equal. Just tried filling out values there, a = 1, b = 1, c = 0 and i arrived at 1 with first one and zero for second one.

Sure its correct?

Delphi91

Use a truth table to evaluate each expression. You will find out that they are not equal.

Or using the laws of Boolean Algebra, the first expression simplifies to ABC(bar) [i.e. bar over the C] while the second evaluates to A(bar)BC(bar) [i.e. bar over the A and the C].

PM me if you want the truth table version.

Mike

Misticles

i got part 1 to equal part 2, got it down to b=b

techguy

Thanks guys, the notes I got 1st time round were wrong..
it should be:
(' = not)
(A'+B'+AC)' = (B'+A'+C)' (not de morgans but another rule..)

Delphi91

hakko1872003 wrote:

Thanks guys, the notes I got 1st time round were wrong..
it should be:
(' = not)
(A'+B'+AC)' = (B'+A'+C)' (not de morgans but another rule..)

Ok, start with:
(A'+B'+C')' = (B'+A'+C)'

Using De Morgan, you get:
A''.B''.C'' = B''.A''.C' (where '' = a double bar)

This simplifies to
ABC = BAC'

or

ABC = ABC'

These are not the same.

Ghost Train

Delphi91 wrote:

Ok, start with:
(A'+B'+C')' = (B'+A'+C)'

Using De Morgan, you get:
A''.B''.C'' = B''.A''.C' (where '' = a double bar)

This simplifies to
ABC = BAC'

or

ABC = ABC'

These are not the same.

I think you left an A out of the left side

Left hand side
(A'+B'+AC)'
AB(AC)'
AB(A'+C')
AA'B+ABC'
0+ABC'
ABC'=right hand side

Delphi91

eolhc wrote:

I think you left an A out of the left side

Left hand side
(A'+B'+AC)'
AB(AC)'
AB(A'+C')
AA'B+ABC'
0+ABC'
ABC'=right hand side

Opps, mea culpa!