Simplifying Boolean expressions

  • #1
polyglot
8
4
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?
 
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  • #2
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?
:welcome:
I'm not sure "factorise" is the correct term. What axioms or theorems do you have at your disposal? The deMorgan laws?
 
  • #3
deMorgan laws, distributive, absorption law etc. I am assuming the usual ones...?
 
  • #4
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?
 
  • #5
If I cannot 'factorise', then I was going to use distributive but not sure how to apply it and then continue with the simplification?
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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?
 
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  • #10
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!
 
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  • #11
You are aboslutly right! Thanks for your help!
 
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