Simplifying Boolean expressions

This is the first time I am doing logic so the question is probably stupid, but could I just factorise (p∨t)∧[(r v¬r] or perhaps you cannot do that in Boolean algebra?

polyglot said:

Homework Statement: I need to simplify the following expression if at all possible and explain the steps
Relevant Equations: [(p∨t)∧r]∨[(p∨t)∧¬r]

This is the first time I am doing logic so the question is probably stupid, but could I just factorise (p∨t)∧[(r v¬r] or perhaps you cannot do that in Boolean algebra?

I'm not sure "factorise" is the correct term. What axioms or theorems do you have at your disposal? The deMorgan laws?

deMorgan laws, distributive, absorption law etc. I am assuming the usual ones...?

polyglot said:

deMorgan laws, distributive, absorption law etc. I am assuming the usual ones...?

If you replace ##p \vee t## by ##q## can you see anything to do?

If I cannot 'factorise', then I was going to use distributive but not sure how to apply it and then continue with the simplification?

polyglot said:

If I cannot 'factorise', then I was going to use distributive but not sure how to apply it and then continue with the simplification?

Show us what you are thinking. Here's some Latex to help (*):
$$\big[(p \vee t) \wedge r \big ] \vee \big [(p∨t)∧\neg r \big ]$$Is factorise a thing in your course notes? Maybe it's terminology I'm not familiar with in this context.

(*) if you reply to my post you can copy and paste the Latex.

I am thinking if the distributive property states that p ∧ ( q ∨ r) = ( p ∧ q) ∨ ( p ∧ r) surely in my original question I just need to apply it in reverse and obtain (p∨t)∧[(r v¬r] . I think this is distributive - factorising does not appear on my notes. Is this correct?

polyglot said:

I am thinking if the distributive property states that p ∧ ( q ∨ r) = ( p ∧ q) ∨ ( p ∧ r) surely in my original question I just need to apply it in reverse and obtain (p∨t)∧[(r v¬r] . I think this is distributive - factorising does not appear on my notes. Is this correct?

That looks like a good first step.

Then, I think I know how to go on: r v¬r is always True, and therefore I will end up with p∨t. Does it look OK to you?

polyglot said:

Then, I think I know how to go on: r v¬r is always True, and therefore I will end up with p∨t. Does it look OK to you?

Yes. The point of studying boolean algebra is to make the obvious difficult, isn't it? Only joking!

You are aboslutly right! Thanks for your help!