Proving a Derived Rule with Natural Deduction: Need Help!

  • Thread starter Thread starter LauraSuh
  • Start date Start date
  • Tags Tags
    Natural
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
LauraSuh
Messages
4
Reaction score
0
Hello smart people!
I was having some troubles proving this derived rule using Natural deduction:

¬(∃y.Q(y) ∧ T(y))
------------------------
∀x.Q(x) → ¬ T(x)

I got stuck in the very first line, because of the "NOT". I can't do anything if I don't take it out of there...

I know that ¬(a ∧ b) = ¬a ∨ ¬b, but I don't know how to prove that as well...

If I could only get past this very first step I'd be able to finish it, but I've been trying for hours and I can't get around it.
Please help!
 
Physics news on Phys.org
To prove DeMorgan's Law, you might want to write out a truth table.
 
It's also helpful to note that:

[itex]p \implies q \equiv \neg p \vee q[/itex]
 
I would use the identity

##\exists x \phi\equiv \neg\forall x\neg\phi## together with the theorem ##\neg\neg\phi\leftrightarrow\phi## and then de Morgan's Law.

If you need more help you will need to specify which Natural Deduction system you are using. There are many. To specify yours you need to list all the rules of inference and replacement rules you have available to you.