Simplifying Boolean expressions

[polyglot]

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?

 

[PeroK]

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?

 

[polyglot]

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

 

[PeroK]

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?

 

[polyglot]

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

 

[PeroK]

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.

 

[polyglot]

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?

 

[PeroK]

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.

 

[polyglot]

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?

 

[PeroK]

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!

 

[polyglot]

You are aboslutly right! Thanks for your help!