I'm struggling to work out a problem I know the answer to. I've got this far:

A'.B'.C' + B

...but I need to further reduce it to

A'.C' + B

What rules should I be applying to make that final step?

Thanks

 

You can do a Karnaugh map simplification for a proven answer.

But think about it this way. If B=1, then the expression is true anyway.
If B=0, then B'=1 and the expression depends only on A'C'.

Does that help?

 

I can't remember the rule name, but if you draw out a quick truth table for the remaining terms, it should be obvious. This is something from experience that I pick out of an expression.

Let Z=A'.C', then your last expression is B'.Z + B. If B=1 then the B'Z term drops out. If B=0, then Z is the result. Hence, the B' is redundant in the first term.

 

Thanks so much for both your answers. This has made it much clearer in my head, I can see exactly why it works now.

If anyone else knows the 'proper' name of the rule that I am applying to arrive at the answer, that would also be helpful (so I can document my work) - but I can see why it works now, which I couldn't quite see before

 

Rather than learning the rules, learn to use a Karnaugh map. It works every time.

 

Unfortunately I've got to prove this works using Boolean Algebra only. I'm not allowed to use Karnaugh maps for this particular problem.

That's why I know the answer (having used Karnaugh to arrive there), but was struggling to see how to get to the same place with Boolean algebra alone.

 

In that case, work backwards:

A'C'+B=
A'C'(B+B')+B=
A'C'B+A'C'B'+B=
B(1+A'C')+A'C'B'

 

Unfortunately I've got to prove this works using Boolean Algebra only. I'm not allowed to use Karnaugh maps for this particular problem.

That's why I know the answer (having used Karnaugh to arrive there), but was struggling to see how to get to the same place with Boolean algebra alone.

So, you used the Karnaugh map incorrectly. You need to wrap the cells around as in a cylinder.

Wrapping A'B'C' with A'BC' reduces to A'C'.

 

Actually he said that he got the answer through a Karnaugh map and it was correct.
But K-maps don't reveal you the progression of Boolean operations to reach the result, which was his problem to begin with.