Simplifying this equation using Boolean Algerbra

[Ese]

Homework Statement

Where
. or SPACE (" ") = AND
+ = OR
` = NOT (Complementary)

Homework Equations

I've been having problems with the second question of my assignment. Here's the equation:
[( a + bc ) + ( b` + a'c )]' = F

The Attempt at a Solution

So far, all I've done is this:
[( a + bc ) + ( b` + a'c )]'
= [(aa+ac+ba+bc) + (b'b' + b'c + a'b' + a'c)]'
= [a+...

Now I'm stuck. Please help me out. :(

 

[ehild]

[( a + bc ) + ( b` + a'c )]' = F

Is it [( a + bc ) + ( b' + a'c )]'?

The Attempt at a Solution

So far, all I've done is this:
[( a + bc ) + ( b` + a'c )]'
= [(aa+ac+ba+bc) + (b'b' + b'c + a'b' + a'c)]'
= [a+...

I do not understand what you did. Try to simplify the expression between the square brackets. Do you know the relation a+a'c=a+c? Do you know De Morgan's laws?

ehild

 

[Ese]

I tried this:

= [(aa+ac+ba+bc) + (b'b' + b'c + a'b' + a'c)]'
= [a+...

[( a + bc ) + ( b` + a'c )]' = F
= (a+bc)' + (b'+a'c)'
= (a'+(bc)') + (b+(a'c)')
= a.b'c' + b'.ac'
= ab'c' + b'ac'
= ab'c'

Is this correct?

 

[shackdaddy836]

One thing you did wrong was that you forgot to change the symbols for the " ' " equations

ex: (a + b)' = a'b'

So, for your first couple lines:

[( a + bc ) + ( b` + a'c )]'

= (a+ bc)'(b'+ a'c)'
= (a'(bc)')(b(a'c)')
= ...