Math Is Fun Forum

Bool

Member

Registered: 2010-03-22

Posts: 3

Boolean Algebra Equivelance

Using the rules of boolean algebra prove (state which rule is used at each step):-

So far I have,

This is where I hit the brick wall.  I've also tried doing the Extended De Morgan rule first but that doesn't seem right to me.

any help appreciated.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Boolean Algebra Equivelance

Before doing any work on proving the statement, make sure you understand it first.  The statement on the left reads:

If p(x) is true, then at least one of q(x) and r(x) is false.

It should be intuitive that this means at least one of {p(x), q(x), r(x)} is false, which is the statement on the right.

Your first statement is wrong.  When you negate a statement "a or b" using DeMorgan, it becomes "not a and not b".  You negated "a or b" to "not a or b".

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Bool

Member

Registered: 2010-03-22

Posts: 3

Re: Boolean Algebra Equivelance

Hi,

Is the first statement correct where I rewrite the implication arrow?

Should the second statement that uses the Extended De Morgan then be:-

White_Owl

Member

Registered: 2010-03-03

Posts: 106

Re: Boolean Algebra Equivelance

To get rid of implication arrow you can use the conversion rule

BTW, I am not sure this rule has a name.
So the expression in the first brackets should be converted:

And the second step is using De Morgans law

Bool

Member

Registered: 2010-03-22

Posts: 3

Re: Boolean Algebra Equivelance

I guess this is a trickier subject than I thought judging by the replies.  Is there anyone out there who can check my final answer to this.  I'm submitting it tomorrow.

Step1
------

Rewriting

Step2
------
De Morgan (ii)

Step3
------
Extended De Morgan (ii)

This is the best I can come up with so it will have to do.  Looks correct to me anyway